09/30/2021
This is the final discussion section before midterm exam and we're going to talk about the last three sections of the lecture.
Chain Rules
If $f$ and $g$ are two differentiable functions so that the composition $f\circ g$ is defined, then the derivative of the composition $f\circ g$ is given by the formula$$(f\circ g(x))'=f'(g(x))g'(x).$$Be careful that here we have $g$ inside $f'$ and we must multiply by a derivative of $g$. Let's see three examples.
Find the derivative of the function $f(t)=\sqrt[3]{1+\tan t}$. Here we can write $g(x)=\sqrt[3]{x}$ and $x(t)=1+\tan t$, then via the chain rule, we have$$f'(t)=g'(x(t))x'(t)=\frac{\frac{1}{\cos^2 t}}{3\sqrt[3]{(1+\tan t)^2}} =\frac{1}{3\cos^2 t+3\sin t\cos t}.$$ Find the derivative of the function $y=\sin\sqrt{1+x^2}$. We can write $y=f\circ g\circ h(x)$ where $f(t)=\sin t$, $g(s)=\sqrt{s}$ and $h(x)=1+x^2$. Then by the chain rule,$$y'=(f\circ g\circ h)'(x)=f'(g\circ h(x))g'(h(x))h'(x)=\left(\cos\sqrt{1+x^2} \right)\frac{x}{\sqrt{1+x^2}} =\frac{x}{\sqrt{1+x^2}}\cos\sqrt{1+x^2}.$$ $y=\left[x^2 +(1-3x)^5\right]^3$. $$y'=3\left[x^2+(1-3x)^5\right]^2\left(2x+5(1-3x)^4(-3)\right) =[6x-45(1-3x)^4][x^2 +(1-3x)^5]^2.$$Implicit Differentiation
This is a way to find derivative of a function given by an equation $F(x,y)=0$, i.e. it's given implicitly. Actually the computation requires some understanding of partial derivative, but it's not too difficult to handle.
Find $\frac{\diff y}{\diff x}$ by implicit differentiation: $\cos (xy)=1+\sin y$. Firstly we take the derivative with respect to $x$ to both sides of the equation and get $$ \frac{\diff}{\diff x} (\cos (xy)-1-\sin y)=0. $$ The derivative of $\cos (xy)$ with respect to $x$ reads $$ \frac{\diff}{\diff x}(\cos (xy))=-\big(\sin (xy)\big)(y+x\frac{\diff y}{\diff x}) $$ by the chain rule, and the derivative of $-1$ is $0$. For the final term, we apply again the chain rule to get $$ \frac{\diff}{\diff x} (-\sin y)=-(\cos y)\frac{\diff y}{\diff x}. $$ Therefore, we get the simplified equation $$ -\big(\sin (xy)\big)(y+x\frac{\diff y}{\diff x})-(\cos y)\frac{\diff y}{\diff x}, $$ and we can simplify the left-hand side to be $$ -(x\sin (xy)+\cos y)\frac{\diff y}{\diff x}-y\sin (xy)=0. $$ This implies that $$ \frac{\diff y}{\diff x} =-\frac{y\sin (xy)}{x\sin (xy) +\cos y}. $$ $y\sin (x^2)=x\sin (y^2)$. Differentiate both sides and we get $$ \frac{\diff y}{\diff x}\sin (x^2) +\big(y\sin (x^2)\big)(2x)-\sin (y^2) -\big(x\cos y^2\big)(2y\frac{\diff y}{\diff x})=0, $$ hence $$ (\sin x^2 -2xy\cos y^2)\frac{\diff y}{\diff x} =\sin y^2 -2xy\sin x^2, $$ and finally $$ \frac{\diff y}{\diff x} =\frac{\sin y^2 -2xy\sin x^2}{\sin x^2 -2xy\cos y^2}. $$ $y\cos x=1+\sin (xy)$. Differentiate both sides and we get $$ \frac{\diff y}{\diff x}\cos x-y\sin x-\big(\cos (xy)\big)\left(y+x\frac{\diff y}{\diff x}\right)=0, $$ hence we have $$ \big(\cos x-x\cos (xy)\big)\frac{\diff y}{\diff x} =y\big(\sin x +\cos (xy)\big), $$ and therefore $$ \frac{\diff y}{\diff x} =\frac{y(\sin x+\cos (xy))}{\cos x-x\cos (xy)}. $$Related Rates
Problems in this section are mostly situation simulation problems. That is, they'll give you a situation and you need to solve this problem by subtracting out data and transcribing everything into mathematics. Let's see two examples.
Each side of the square is increasing at a rate of $6$cm/s. At what rate is the area of the square increasing when the area of the squre is $16$cm2? Assume the side of the square is the function $s(t)$, depending on the time $t$, and the area of the square is $A(t)=s(t)^2$. What we want is the value $A'(t)$ when $A(t)=16$cm2. Note that $$ A'(t)=\big((s(t)^2\big)'=2s(t)s'(t), $$ hence what we want is THE value of $s(t)$ at the time $t$ when $A(t)=16$cm2. If $A(t)=16$cm2, then by the relation $A(t)=s(t)^2$ we know that $s(t)=4$cm, and the problem tells us that $s'(t)=6$cm/s, hence we conclude that $$ A'(t)=2\times 4\mathrm{cm}\times 6\mathrm{cm}/\mathrm{s}=24\mathrm{cm}^2/\mathrm{s}. $$ If two resistors with resistence $R_1$ and $R_2$ are connected in parallel, then the total resistence $R$, measured in ohms($\Omega$), is given by $$ \frac{1}{R} =\frac{1}{R_1} +\frac{1}{R_2}. $$ If $R_1$ and $R_2$ are increasing at rates $0.3\Omega$/s and $0.2\Omega$/s respectively, how fast is $R$ changing when $R_1=80\Omega$ and $R_2=100\Omega$?