diff --git a/北京师范大学-香港浸会大学联合国际学院/index.tm b/北京师范大学-香港浸会大学联合国际学院/index.tm index a90e0f2..a49283f 100644 --- a/北京师范大学-香港浸会大学联合国际学院/index.tm +++ b/北京师范大学-香港浸会大学联合国际学院/index.tm @@ -1,4 +1,4 @@ - + > @@ -20,12 +20,16 @@ + + <\itemize> + + diff --git a/北京师范大学-香港浸会大学联合国际学院/常微分方程/Chapter_2.tm b/北京师范大学-香港浸会大学联合国际学院/常微分方程/Chapter_2.tm new file mode 100644 index 0000000..1fa69fd --- /dev/null +++ b/北京师范大学-香港浸会大学联合国际学院/常微分方程/Chapter_2.tm @@ -0,0 +1,1177 @@ + + +> + +<\body> + + |>>> + + + + General form\ + + <\equation*> + F|)>=0 + + + <\example> + + + The motion of the object is governed by the Newton's law. Let + > be the velocity of the object at time . + + <\equation*> + m*v=m*g-\*v, + + + where is the mass of the object, is the gravitational + constant, and > is the coefficient of air resistant + force. + + This is a first order linear ODE. + + + + + <\example> + Find the general solutions of + + <\equation*> + u+u=2 + + + idea: combine +u> into the derivative of another + function.\ + + Consider the product rule of differentiation: + + <\equation*> + =f*g+f*g. + + + Let ,g=e>, then + + <\equation*> + *e|]>=u*e+u*e=e*+u|)>. + + + Now, we multiply the original equation by >: + + <\equation*> + +u|)>*e=2*e\*e|]>=2*e\u*e=2*e*\t=2*e+c + + + + Divide by >: + + <\equation*> + u=e*+c|]>=2+c*e. + + + + + Consider the 1st order linear ODE (standard form): + + <\equation*> + +p*u=q.> + + + Multiply left side by >: + + <\equation*> + +p*u|]>*\=u*\+p*u*\. + + + We want + + <\equation*> + \=g,p*\=g + + + i.e. + + <\equation*> + \=p**\\=ep*\t>>. + + + Check (by the chain rule and fundamental theorem of calculus) + + <\equation*> + \=ep*\t>*p=\*p. + + + Derivation directly: + + <\equation*> + \=p**\\|\>=p\|]>=p + + + <\equation*> + \ln*\=p*\t\\=ep*\t.> + + + So the original ODE becomes + + <\equation*> + *u|]>=\*q\\*u=\*q*\t+c + + + <\equation*> + \=>*\*q*\t+c|]>>, + + + where + + <\equation*> + =ep*\t>> + + + is called the . + + <\example> + Find the of + + <\equation*> + |)>*+2*t*y=4*t + + + Rewrite it in the standard form + + <\equation*> + y+>*y=>. + + + Here =>>. Then find the + integrating factor >: + + <\equation*> + \=ep*\t>=e|)>>=4+t. + + + So the solution is + + <\equation*> + y=>*|)>*>*\t+c|)>=>*+c|)> + + + + <\example> + Solve the + + +2*y=4*t>>|y=2>>>>> + + First rewrite the equation into the standard form: + + <\equation*> + y+*y=4*t. + + + Find the integrating factor: + + <\equation*> + \=e*\t>=e>=e>=t. + + + The general solution is + + <\equation*> + y=>* + t*4*t*\t+c|]>=>*+c|]>=t+>. + + + Plugging the initial condition: + + <\equation*> + y=2\1+=2\. + + + The solution of the initial value problem is + + <\equation*> + y=t+>. + + + + \; + + + + <\example> + Solve + + <\equation*> + =|1-y>. + + + This equation is nonlinear. We can separate the variables and + as follows + + <\equation*> + |)>*\y=x*\x\|)>*\y=x*\x + + + <\equation*> + \y-|3>=|3>+c + + + This is an example of . + + + <\definition> + An ODE in the form of + + <\equation*> + y|\x>=|g> + + + is called + + + We can solve it as follows + + <\equation*> + f*\x=g*\y + + + <\example> + Solve the initial value problem + + <\equation*> + =+4*x+2|2*>,y=-1 + + + \ + + <\equation*> + +4*x+2|)>*\x=2**\y + + + <\equation*> + \x+2*x+2*x=y-2*y+c. + + + Plugging the initial condition + + <\equation*> + y=-1\0=1+2+c\c=-3. + + + The solution to the initial value problem is + + <\equation*> + x+2*x+2*x=y-2*y-3. + + + <\question> + What is the domain and range of the solution? + + + + <\example> + Recall the differential equation for continuous compound interests: + + <\equation*> + u=r*u. + + + Note this equation is both linear and separable. As a separable + equation, we have + + <\equation*> + u|\t>=r*u\u|u>=r*\t\u|u>=r*\t\ln + =r*t+c + + + <\equation*> + \=c*e\u=c*e. + + + If the initial condition is =u>,then we + find >. + + <\exercise> + Solve the equation by the method of integrating factor. + + + + + + + + Suppose =c> is an solution of some ODE. + Taking x> on both sides of the solution. + + <\equation*> + x>*\=x>*c\\|\x>>+\|\y>>*y|\x>=0\>+>*y=0, + + + where + + <\equation*> + M=\\,N=\*\. + + + \; + + <\example> + Solve >+*y=0>. + + <\answer*> + Guess the solution. Let =x+y*x>. Then + + <\equation*> + \=2*x+y,\=2*x*y. + + + So + + <\equation*> + 0=\+\*y=x>*\ + + + So the solution is + + <\equation*> + \=c. + + + + + <\definition*> + An ODE of the form + + <\equation*> + M+N*y=0M*\x+N*\y=0 + + + is called if there exists + > such that + + <\equation*> + \=M,\=N. + + + The solution of the equation is + + <\equation*> + \=c, + + + where is an arbitrary constant. + + + + + <\theorem> + Suppose an ODE can be written in the form + + <\equation> + M+N*y=0M*\x+N*\y=0 + + + where the functions > and > are all + continuous in the rectangular region + \>. Then Eq. (1) is + an exact differential equation + + <\equation*> + M=N,\\R. + + + + <\proof> + \>. Suppose Eq. (1) is + exact. Then there exists a > such that + + <\equation*> + \=M,\=N. + + + Then + + <\equation*> + M=\,N=\. + + + Since ,N> are continuous, we have + > and > are continuous. + So\ + + <\equation*> + \=\. + + + i.e. + + <\equation*> + M=N. + + + \> Suppose + =N>. We want to find a function + > such that =M> and + =N>. Let + + <\equation*> + \=M*\x+h. + + + Then =M>, and + + <\equation*> + \=\M*\x+h. + + + We want =N>, that is + + <\equation*> + h=N-\M*\x. + + + We need the RHS to be independent of . That is + + <\equation*> + |\x> -\ + *M*\x|]>=0. + + + Let's check: + + <\eqnarray*> + |\x>-\*M*\x|]>>||-\ + \M*\x=N-M=0.>>>> + + + \; + + + \; + + <\example> + Solve the ODE + + <\equation*> + |)>+*e-1|)>*y=0. + + + \ + + <\eqnarray*> + >||>>|>||>>>> + + + So =N>, and the equation is exact. + + Next, let + + <\equation*> + \=M d*x=y*+2*x*e + d*x=y*+x*e+h. + + + Then + + <\equation*> + \=sin x+x*e+h=N=sin + x+x*e-1 + + + <\equation*> + \h=-1\h=-y. + + + So the solution is + + <\equation*> + \=y*+x*e-y=c. + + + + <\exercise*> + Solve the above equation, but using =N + d*y+h> first. + + + <\question*> + What is the relationship between separable and exact equations? + + + + + Sometimes we can multiply a function to a non-exact equation to make it + exact. Take a function \0>,\ + + <\eqnarray*> + d*x+N + d*y>||>| + d*x+N d*y|]>>||>|*M + d*x+\*N + d*y>||>| + d*x+ d*y>||>>> + + + where =\*M,=\*N>. + Then let + + <\equation*> + =\*M+\*M,=\*N+\*N. + + + We want =>, i.e. + + <\equation*> + \*M+\*M=\*N+\*N. + + + Let's choose > such that =0>. Then the + above equation reduces to + + <\equation*> + \*M=\*N+\*N\\=-N|N>*\. + + + If the function -N|)>/N> is a function + of only, then we can solve > as a separable + equation. Here > is called an integrating factor. + + <\wide-block> + + <\example> + Solve the ODE + + <\equation*> + |)>++x*y|)>*y=0. + + + It's first order, nonlinear, and not separable. + Check if it's exact: + + <\equation*> + M=3*x+2*y,N=2*x+y. + + + Not exact!. Next, try integrating factors. + + <\equation*> + -N|N>=+x*y>= + + + is a function of only! Let + + <\equation*> + \=*\\\=x. + + + Then multiply > to the original equation: + + <\equation*> + x*|)>+x*+x*y|)>*y=0 + + + <\equation*> + *y+x*y|)>++x*y|)>*y=0 + + + Double check the new equation is exact! Then solve it as usual + (exercise). + + >>> + + + Similarly, if -M|)>/M> is a function of + only, then we can use the integrating factor + > solving + + <\equation*> + \=-M|M>*\. + + + + + Consider the first order ODE: + + <\equation*> + y=f + + + Draw small arrows as a vector |)>> + at many points > + + Online plotter: + + + + <\example*> + Consider + + <\equation*> + y=> + + + + \; + + + + + + <\theorem*> + Consider the initial value problem + + <\equation*> + y+p*y=q,y|)>=y. + + + If are continuous on an interval > + containing >, then the IVP has a unique solution on + . + + + <\example*> + Consider + + <\equation*> + t*y+2*y=4*t,y=2. + + + Solve it by integrating factors,\ + + <\equation*> + y+*y=4*t\\=exp + d*t|]>=t. + + + <\equation*> + y=>4*t + d*t+c|]>=>+c|]>=t+>. + + + Plugging =2>, we obtain . The solution + is + + <\equation*> + y=t+>. + + + Now, =,q=4*t>. So + are continuous in ,0|)>\|)>>. + But |)>> only, so we know from the + theorem the IVP has a unique solution in + |)>>, which is + + <\equation*> + y=t+>,t\|)>. + + + If the initial condition is changed to =2>, then + the solution is + + <\equation*> + y=t+>,t\,0|)>. + + + + + + <\theorem*> + Consider the initial value problem + + <\equation*> + y=f,y|)>=y. + + + If and f> are continuous on a + rectangular domain \> + containing the point ,y|)>>. Then the + IVP has a unique solution in some interval containing + >. + + + <\example*> + Consider the IVP. + + <\equation*> + =+4*x+2|2*>,y=-1 + + + It is separable. Let's solve it first,\ + + <\equation*> + 2*d*y=+4*x+2|)>*d*x\y-2*y=x+2*x+2*x+c + + + <\equation*> + y=-1\c=3. + + + The solution is + + <\equation*> + y-2*y=x+2*x+2*x+3 + + + <\equation*> + y=+2*x+2*x+3|)>>|2>=1\+2*x+2*x+4>=1-+2*x+2*x+4>. + + + Here and f> are continuous + everywhere except . + + + <\example*> + Consider + + <\equation*> + y=y,y=0*0|)> + + + First, let's solve it as a separable equation. + + <\equation*> + y d*y=d*t\*y=t+c + + + Plugging =0> yields . So + + <\equation*> + y=\*t|)> + + + are two solutions. In addition + + <\equation*> + y=0 + + + is also a solution. In fact, we have infinitely many solutions defined + as + + <\equation*> + y=|t>>||*>*t-t|)>|]>,>|t>>>>>,y=|t>>|*|)>|]>,>|t>>>>> + + + for any \0>. ( check is + continuous and differentiable at >.) + + |gr-frame|>|gr-geometry||gr-color|red|gr-auto-crop|true|||>>|||>>||>>|||>>||>>|||>>||>>|||>>||>>||>>|||>>|>>> + + + In fact,\ + + <\equation*> + f=y,\ f=*y. + + + So f> is discontinuous near + >. So there exists no rectangle + containing > such that + f> are both continuous in . So we can't gaurantee the existence + and uniqueness of solution for the IVP. + + <\note> + One may not be able to find all solutions to nonlinear equations using + one method.\ + + + <\example*> + Consider + + <\equation*> + =y,y=y\0 + + + We can first solve it as a separable eqn. + + <\equation*> + y*d*y=d*t\-y=t+c\y=-\y=->>. + + + Now ,\ f=2*y> are continuous + everywhere. However, the solution is not defined for every . + For example, if \0>,then the solution is defined + only in ,>|)>>. + + + + + + + <\equation*> + m*v=m*g-\*v, + + + where is the velocity, > are constants.\ + + <\itemize> + Analyze the solutions using direction field. + + Solve it by integrating factors. + + <\equation*> + v+|m>*v=g + + + integrating factor + + <\equation*> + \=e\/m>=e|m>*t> + + + <\equation*> + v=e|m>*t>*g*e|m> + t>*d*t+c|]>=e|m>*t>*>*e|m>*t>+c|]>=>+c*e|m>*t>. + + + If the initial condition is =v>. Then + -g*m/\>. So the solution of the IVP is + + <\equation*> + v=>+->|]>*e|m>*t>. + + + So + + <\equation*> + lim\> + v=>. + + + All other solutions converge to the + > as \>. This + equilibrium solution is a one. + + + + + Assume the annual interest rate is . The continuous rate of + deposit/withdrawal is . Then the ODE model for the total balance + > is + + <\equation*> + u=r*u+k. + + + integrating factor + + <\equation*> + \=e + + + <\equation*> + u=e*k*e+c|]>=e**e+c|]>=-+c*e. + + + If the initial condition is =u>, then + +k/r>. So the solution of the IVP is + + <\equation*> + u=-++|)>*e. + + + The equilibrium solution is >, and it is an + one since all other solutions diverge from it as + \>. + + + + + + <\equation*> + y=r*y, + + + The solution is + + <\equation*> + y=y*e + + + where =y>.\ + + <\itemize> + If 0>, we have exponential growth + + If 0>, we have exponential decay, such as + radioactive decay. + + + + + <\equation*> + y=*y. + + + Note that the right-hand-side depends on only. In general, ODE + of the form + + <\equation*> + y=f + + + is called . There are two equilibrium solutions + + <\equation*> + y=0,y=. + + + From the direction field we can tell the equilibrium solution + is unstable, while the solution > is stable. + + <\wide-centered> + |png>|0.4par|||> + + + Now let's solve the equation: + + <\equation*> + *y>=d*t\*y>=d*t\ + + + <\equation*> + *y>=+=+B*y|y*>\1=A*+B*y + + + <\equation*> + y=0\A=1/r,y=r/a\B=a/r + + + So + + <\equation*> + *y>=+*d*y=*ln + +|)>*ln + =*ln -*ln + + + + <\equation*> + =*ln |>=d*t=t+c + + + <\equation*> + \|>=e>=c*e\=c*e. + + + <\equation*> + \y=|1+a*c*e>=+a*c>=*e+a>. + + + Suppose the initial condition is =y>, then\ + + <\equation*> + c=|r-a*y>\y=|y>*e+a>=|a*y+|)>*e>=|y+|)>*e>>>>>>, + + + where >. Note that =*y=r*|)>*y> + + <\enumerate> + If y\K>, then + > is an increasing function, and + \> y=K>, but + \K> for all 0>. Moreover, + \> y=0>, + and + + <\equation*> + y= = + f=f*=f*f, + + + where + + <\equation*> + f=r*|)>*y,f=r*|)> + + + <\enumerate> + If y\>, then + \0>, so the graph is concave up. + + If \y\K>, then + \0>, so the graph is concave down. + + + If \K>, then > is an + decreasing function, and \> + y=K>, but \K> for all + 0>. Moreover, \> + y=0>, and \0> + for all . + + + + + Consider a general 1st order ODE + + <\equation*> + y=f. + + + Take ,y|)>>, then + + <\equation*> + y|)>=f,y|)> + + + <\equation*> + y|)>=lim0> + +h|)>-y|)>|h>\|)>-y|)>|t-t> + + + if -t|\|>> is small. So + + <\equation*> + y|)>\y|)>+-t|)>*y|)>=y|)>+-t|)>*f,y|)> + + + Let + + <\equation*> + y=y+-t|)>*f,y|)> + + + So \y|)>>. Repeat this + process, we obtain an algorithm: For a sequence of + ,t,t,\> + + <\equation*> + =y+-t|)>*f,y|)>>>>>> + + + This sequence of ,y,y,\> is an + approximation of the true values |)>,y|)>,y|)>,\> + + \; + + \; + + \; + > + + +<\initial> + <\collection> + > + + + + + + + + + + + +<\references> + <\collection> + > + > + > + > + > + > + > + > + > + > + > + > + > + > + > + > + > + > + + + +<\auxiliary> + <\collection> + <\associate|toc> + |math-font-series||font-size||Chapter + 2: First Order Differential Equations> + |.>>>>|> + + + |math-font-series||1Method + of integrating factors> |.>>>>|> + + + |math-font-series||2Separable + equations> |.>>>>|> + + + |math-font-series||3Exact + Equations> |.>>>>|> + + + |3.1Motivation and definition + |.>>>>|> + > + + |3.2Theorem and method + |.>>>>|> + > + + |3.3Integrating factors + |.>>>>|> + > + + |math-font-series||4Direction + fields> |.>>>>|> + + + |math-font-series||5The + Existence and Uniqueness Theorem> |.>>>>|> + + + |5.1Linear equations + |.>>>>|> + > + + |5.2Nonlinear equations + |.>>>>|> + > + + |math-font-series||6Applications> + |.>>>>|> + + + |6.1Falling object in the air + |.>>>>|> + > + + |6.2Compound interest with + deposits/withdrawals |.>>>>|> + > + + |6.3Population dynamics + |.>>>>|> + > + + |6.3.1Exponential growth + |.>>>>|> + > + + |6.3.2Logistic growth + |.>>>>|> + > + + |math-font-series||7Euler's + method> |.>>>>|> + + + + \ No newline at end of file diff --git a/北京师范大学-香港浸会大学联合国际学院/微积分/Chapter_3.tm b/北京师范大学-香港浸会大学联合国际学院/微积分/Chapter_3.tm new file mode 100644 index 0000000..f472754 --- /dev/null +++ b/北京师范大学-香港浸会大学联合国际学院/微积分/Chapter_3.tm @@ -0,0 +1,1101 @@ + + +> + +<\body> + <\hide-preamble> + >>>>>> + + + + >> + > + > + + + + >> + > + > + + >>>>> + + >>>> + + |-0.3em|>|0em||0em|>>>> + + + <\slideshow> + <\slide> + > + + <\table-of-contents|toc> + Derivatives + of Polynomials and Exponential Functions> + |.>>>>|> + + + Power functions + |.>>>>|> + > + + Linear combination + |.>>>>|> + > + + Exponential functions + |.>>>>|> + > + + The + Product and Quotient Rules> |.>>>>|> + + + The product rule + |.>>>>|> + > + + The quotient rule + |.>>>>|> + > + + Derivatives + of Trigonometric Functions> |.>>>>|> + + + >>> |.>>>>|> + > + + The + Chain Rule> |.>>>>|> + + + Implicit + differentiation> |.>>>>|> + + + Derivatives + of Logarithmic Functions> |.>>>>|> + + + Logarithmic differentiation + |.>>>>|> + > + + Rates + of Change in the Economics and Social Sciences> + |.>>>>|> + + + skip> + |.>>>>|> + + + skip> + |.>>>>|> + + + Linear + Approximations and Differentials> + |.>>>>|> + + + Linear approximations + |.>>>>|> + > + + Differentials + |.>>>>|> + > + + + + + + + <\itemize> + If =c>, where is a constant. + Then + + <\equation*> + f=lim0> + -f|h>=lim0> + =0. + + + If =x>, then + + <\equation*> + f=lim0> + -f|h>=lim0> + -x|h>=1. + + + If =x>, then + + <\equation*> + f=lim0> + -f|h>=lim0> + -x|h> + + + <\equation*> + =lim0> |h>=lim0> + =2*x. + + + In general, if =x,n\\>, + then + + <\equation*> + f=n*x. + + + Hint: use the binomial expansion + + <\equation*> + =a+n*a*b+|2>*a*b+\+b=>|>>>>*a*b, + + + where + + <\equation*> + >|>>>>=!>. + + + The above results can be generalized to + =x,a\\>.\ + + <\equation*> + =a*x>. + + + + \; + + From this result we can find the derivative of any polynomial. + + + + <\equation*> + *f+c*g|]>=c*f+c*g>. + + + <\example> + If =x+5*x+6*x-12,>then + + <\equation*> + f=100*x+15*x+6 + + + + + + Let =b,b\0>. Then + + <\eqnarray*> + >||0> + -f|h>>>|||0> + -b|h>=lim0> + *-1|)>|h>>>|||*lim0> + -1|h>.>>>> + + + In fact,\ + + <\equation*> + lim0> -1|h>=f + + + if it exists, and + + <\equation*> + f=f*b. + + + In fact,\ + + <\equation*> + |)>=*b>. + + + In particular, + + <\equation*> + |)>=e>. + + + <\question> + Prove the limit exists. Hint: tranform the limit to the form + containing the term \>|)>>. + + + + to see a proof that the limit of the sequence + |)>>exists as + \>. + + \; + + <\definition> + The of a function + > is + + <\equation*> + |f>, + + + and the is + + <\equation*> + 100*|f>. + + + + \; + + + + + + \; + + Motivation: How to find the derivative function of + =x*e>? + + \; + + <\theorem> + If and are both differentiable + at , then is differentiable at . Moreover, + + <\equation*> + =f*g+f*g. + + + + <\proof> + From the definition, + + <\eqnarray*> + >||0> + *g-f*g|h>>>|||0> + *g*g+f*g>-f*g|h>>>|||0> + *-f|]>+f*-g|]>|h>>>|||0> + *-f|]>|h>+lim0> + *-g|]>|h>>>|||0> + g*lim0> + -f|h>+lim0> + f*lim0> + -g|h>>>|||*f+f*g.>>>> + + + \; + + + Intuitive explanation + + <\wide-block> + || + |png>|0.33par|||> + |<\cell> + Let ,v=g>. For small change + x>, let u,\v> be the change + in , respectively. Let > + be the change in . + + <\eqnarray*> + >||u|)>*v|)>-u*v>>|||v+v*\u+\u*\v-u*v>>|||v+v*\u+\u*\v>>>> + + + <\eqnarray*> + |\x>>||v+v*\u+\u*\v|\x>>>|||v|\x>+v*u|\x>+\u*v|\x>>>||>|+v*u+0as\x\0>>|||+v*u>>>> + + >>> + + + \; + + <\example> + \; + + <\enumerate-alpha> + If =x*e>, find + >.\ + + <\equation*> + |)>=x*e+x*|)>=e+x*e=e*. + + + Find the th derivative, + >>. + + <\eqnarray*> + |)>>||*|]>>>|||+e=e*.>>>> + + + Repeat this process one more time, we get\ + + <\equation*> + |)>=e*. + + + In general + + <\equation*> + f>=e*. + + + <\exercise> + Prove the general result using mathematical induction. + + + + + + + <\theorem> + If and are both differentiable + at and \0>, then is + differentiable at . Moreover, + + <\equation*> + |)>=*g-f*g|g>. + + + + \; + + <\exercise> + Prove the quotient rule using the definition of derivative. + + + <\example> + Let +x-2|x+6>>. Then + + <\eqnarray*> + >||+x-2|)>*+6|)>-+x-2|)>*+6|)>|+6|)>>>>|||*+6|)>-+x-2|)>*|)>|+6|)>>>>|||+x+12*x+6-+3*x-6*x|)>|+6|)>>>>|||-2*x+6*x+12*x+6|+6|)>>.>>>> + + + + \; + + + + > + + By the definition + + <\eqnarray*> + >||0> + -sin|h>>>|||0> + *cos+cos*sin-sin|h>>>|||0> + *-1|]>+cos*sin|h>>>|||0> + *-1|]>|h>+lim0> + *sin|h>>>|||*lim0> + -1|h>+cos*lim0> + |h>>>|||.>>>> + + + <\wide-block> + || + |gr-frame|>|gr-geometry||gr-grid||1>|gr-grid-old||1>|gr-edit-grid-aspect|||>|gr-edit-grid||1>|gr-edit-grid-old||1>|gr-grid-aspect|||>|gr-grid-aspect-props|||>|gr-snap||gr-auto-crop|true|gr-snap-distance|5px|||>||>|>|>|>|>||>||>||>|>>> + |<\cell> + First, assume h\|2>> + + <\equation*> + O*A*B|\|>\>|\|>\O*A*D|\|> + + + <\equation*> + *sin\*h\*tan + + + <\equation*> + \*sin\h\tan + + + <\equation*> + \1\\ + + + <\equation*> + \cos h\\1 + + + <\equation*> + \lim0> cos + h\lim0> \lim0> 1 + + + <\equation*> + \1\lim0> + \1. + + + By the squeeze theorem, we have + + <\equation*> + lim0> =1. + + + <\equation*> + \; + + >>> + + + <\eqnarray*> + 0> + -1|h>>||0> + -1|h>*+1|cos+1>>>|||0> + -1|h>*>>|||0> + h|h>*>>|||0> + h|h>**lim0> >>|||*lim0> + h|h>=-*lim0> + *lim0> sin h=0.>>>> + + + For your record + + <\equation*> + 0> =1,lim0> =0>. + + + <\theorem> + and > + + <\equation*> + =cos x,=-sin x. + + + + \; + + Using the quotient rule, we can find the derivatives of other + trigonometric functions. + + <\equation*> + =|)>=*cos-sin*|cos x>= x>=sec x. + + + <\exercise> + Find the derivative of . + + + + + Motivation: If =>>, then + =?> + + \; + + <\theorem> + Suppose is differentiable at + and is differentiable at >. Then + g> is differentiable at , and + + <\equation*> + g|)>=f|)>*g. + + + Another way to say this. Suppose is a function of , + which is differentiable at , and is a function of + , which is differentiable at >, then + is a function of , which is differentiable at + , and + + <\equation*> + z|dx>=z|dy>*y|dx> + + + + <\proof> + By the definition, + + <\eqnarray*> + g|)>>||0> + |)>-f|)>|h>>>|||0> + |)>-f|)>|h>*-g|g-g>>>|||0> + |)>-f|)>|g-g>**lim0> + *-g|h>>>||||)>*g.>>>> + + + \; + + + <\example> + Find > if + =>>. + + + g>, where + =,g=1+x>. + + <\equation*> + F=f|)>*g=>>*2*x=>>. + + + \; + + <\example> + Differentiate -1|)>>. + + + \ + + <\equation*> + y=100*-1|)>*|)>=300*x*-1|)>. + + + \; + + : For example,\ + + <\equation*> + g\h|)>=f|)>|)>*g|)>*h. + + + <\example> + Find > if + =2>> + + <\equation*> + f=*2>*>*. + + + + + + Motivation: How to find > if + =sin y>? + + <\example> + Find |y|dx>|\|>> + if +y=25>. + + + |gr-frame|>|gr-geometry||gr-grid||1>|gr-grid-old||1>|gr-edit-grid-aspect|||>|gr-edit-grid||1>|gr-edit-grid-old||1>|gr-auto-crop|true||>||>|||>||||>> + + Taking x>> on both sides + of the equation (suppose is a function of ): + + <\equation*> + x>*+y|)>=x>* + + + <\equation*> + x>*x+x>*y=0 + + + <\equation*> + 2*x+2*y*y|dx>>=0 + \y|dx>=-=-. + + + For this example, we can solve for and double check + the answer. + + <\equation*> + y=\>=> + near** + + + <\equation*> + y|dx>=>>=->>=-. + + + \; + + <\example> + Lemniscate curve: + + <\wide-tabular> + | + <\script-output|asymptote|default> + % -width 0.5par + + import contour; + + size(10cm, 0); + + draw((-3,0)--(3,0), arrow=Arrow, gray+linewidth(0.5pt)); + + label("$x$", (3,0), align=E); + + draw((0,-1.5)--(0,1.5), arrow=Arrow, gray+linewidth(0.5pt)); + + label("$y$", (0,1.5), align=N); + + real f(real x, real y) {return (x^2+y^2)^2-4*(x^2-y^2);} + + guide[][] thegraphs = contour(f, a=(-2,-2), b=(2,2), new real[] + {0}); + + draw(thegraphs[0], linewidth(1pt)); + + label("$(x^2+y^2)^2=4(x^2-y^2)$", (0,1), align=E); + |ps>|0.5par|||>> + + \; + >>> + + + <\equation*> + x>*+y|)>=x> + -y|)>|]> + + + <\equation*> + 2*+y|)>*|)>=4*|)> + + + <\equation*> + \y=+y|)>|y*+y|)>+2*y> + + + + + + <\eqnarray*> + || + x>>|||>>|x> + x>||x> + b>>|||b*y>>|y>||b>>>|||*x>>>>> + + + <\theorem> + + + <\equation*> + x|)>=*x>,=. + + + + \; + + + + Using the properties + + <\equation*> + ln =ln x+ln y,ln x=y*ln x + + + and implicit differentiation, one can simplify the differentiation of + some functions. + + <\example> + Find y|dx>> if + +1|)>*|+sin + x+2>>>. + + + First, take on both sides. + + <\eqnarray*> + ||+1|)>*|+sin + x+2>>>>|||+1|)>+3*ln + -*ln +sin x+2|)>>>>> + + + Then take x>> on both sides, + + \; + + <\eqnarray*> + *y>||+1>\2**x+-\+sin x+2>>>|||+1>+-+sin x+2|)>>>>|>||=y*+1>+-+sin x+2|)>>|)>>>|||+1|)>*|+sin + x+2>>+1>+-+sin x+2|)>>|)>>>>> + + + <\remark> + Let , then + + <\equation*> + x>=y>*y|dx>=*y. + + + + <\example> + =x>. + + + \ + + <\eqnarray*> + ||>|x> >||x> >>||y>>||=1+ln x>>|>||*>>>> + + + + + Read the slides. + + \; + + <\section> + skip + + + + + + + + + \; + + Suppose is differentiable at , then + + <\equation*> + f=lima> + -f|x-a>\-f|x-a>x a + + + <\equation*> + \f-f\f* + + + <\equation*> + \f+f*>. + + + is called the of >. + And the linear function + + <\equation*> + L=f+f* + + + is called the of >. + + \; + + <\wide-block> + |||||| + |png>|0.4par|||>||gr-color||gr-frame|>||>>>>|0cm> + |<\cell> + <\example> + Estimate the value of > and > + using a linear approximation. + + + Let =>. Then + =>> + + <\equation*> + f\f+f*0.1=2+=2.025. + + + <\equation*> + f\f+f*=2-=1.9975 + + + \; + >>> + + + \; + + <\example> + Approximate the value of >. + + + Let =sin x>, then + =cos x>,\ + + <\equation*> + f\f+f*0.2=0+1\0.2=0.2 + + + <\session|python|default> + <\output> + Python 3.9.6 [/Library/Developer/CommandLineTools/usr/bin/python3]\ + + Python plugin for TeXmacs. + + Please see the documentation in Help -\ Plugins -\ + Python + + + <\unfolded-io> + \\\\ + <|unfolded-io> + 4.1**0.5 + <|unfolded-io> + 2.0248456731316584 + + + <\unfolded-io> + \\\\ + <|unfolded-io> + 3.99**0.5 + <|unfolded-io> + 1.997498435543818 + + + <\unfolded-io> + \\\\ + <|unfolded-io> + math.sin(0.2) + <|unfolded-io> + 0.19866933079506122 + + + <\input> + \\\\ + <|input> + \; + + + + \; + + <\example> + Find an approximate value of >. + + + Let =>. Then + =*x>>.\ + + <\equation*> + f\f*+f*=3+=. + + + <\session|python|default> + <\unfolded-io> + \\\\ + <|unfolded-io> + 29**(1/3) + <|unfolded-io> + 3.072316825685847 + + + <\unfolded-io> + \\\\ + <|unfolded-io> + 83/27 + <|unfolded-io> + 3.074074074074074 + + + + + + Suppose > is differentiable at , then + + <\equation*> + y|dx>=f\dy=f*dx, + + + where we call x,dy> + . + + <\example> + The radius of a sphere was measured and found to be + cm>> with a possible error in + measurement of at most cm>>. Estimate + the maximum error in using this value of the radius to compute the + volume of the sphere? + + + Let denote the radius and volume. Then + + <\equation*> + V=*\*r. + + + Then =4*\*r> + + <\equation*> + d V=V*d*r=4*\*r*d*r + + + Then + + <\equation*> + =4*\*r*\4*\*21*0.05\277*cm. + + + The relative change of the volume is + + <\equation*> + |\|>=\\0.007=7%. + + + + + +<\initial> + <\collection> + > + + + + + + > + + + + + + > + + + + + > + + + + + + + + + + + + + + +<\references> + <\collection> + > + > + > + > + > + > + > + > + > + > + > + > + > + > + > + > + > + > + > + + + +<\auxiliary> + <\collection> + <\associate|toc> + |math-font-series||1Derivatives + of Polynomials and Exponential Functions> + |.>>>>|> + + + |1.1Power functions + |.>>>>|> + > + + |1.2Linear combination + |.>>>>|> + > + + |1.3Exponential functions + |.>>>>|> + > + + |math-font-series||2The + Product and Quotient Rules> |.>>>>|> + + + |2.1The product rule + |.>>>>|> + > + + |2.2The quotient rule + |.>>>>|> + > + + |math-font-series||3Derivatives + of Trigonometric Functions> |.>>>>|> + + + |Derivative of + |font-family|||||sin + x>>>> |.>>>>|> + > + + |math-font-series||4The + Chain Rule> |.>>>>|> + + + |math-font-series||5Implicit + differentiation> |.>>>>|> + + + |math-font-series||6Derivatives + of Logarithmic Functions> |.>>>>|> + + + |6.1Logarithmic differentiation + |.>>>>|> + > + + |math-font-series||7Rates + of Change in the Economics and Social Sciences> + |.>>>>|> + + + |math-font-series||8skip> + |.>>>>|> + + + |math-font-series||9skip> + |.>>>>|> + + + |math-font-series||10Linear + Approximations and Differentials> |.>>>>|> + + + |10.1Linear approximations + |.>>>>|> + > + + |10.2Differentials + |.>>>>|> + > + + + \ No newline at end of file