AP Calculus BC - Chain Rule and Implicit Differentiation

$g(x) = e ^{x^{2}+x}$

Evaluate .

$7 e^{12}$

$12 e^{12}$

$12 e^{11}$

$7 e^{9}$

$e^{7}$

Explanation

To find , substitute $u = x ^{2} + x$ and use the chain rule:

$g(x) = e ^{x^{2}+x}$

$g'(x) = \frac{\mathrm{d} }{\mathrm{d} x} e ^{x^{2}+x}$

$= \frac{\mathrm{d} }{\mathrm{d} x} e ^{u}$

$= \frac{\mathrm{d}u }{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u}e ^{u}$

$= \frac{\mathrm{d}u }{\mathrm{d} x} \cdot e ^{u}$

$= \frac{\mathrm{d} }{\mathrm{d} x}\left ( x ^{2} + x \right ) \cdot \frac{\mathrm{d} }{\mathrm{d} u}e ^{u}$

$= \left ( 2x + 1 \right ) \cdot e ^{u}$

$= \left ( 2x + 1 \right ) \cdot e ^{ x ^{2} + x}$

$g'(x)= \left ( 2x + 1 \right ) e ^{ x ^{2} + x}$

Plug in 3:

$g'(3)= \left ( 2 \cdot 3 + 1 \right ) \cdot e ^{ 3 ^{2} + 3}$

$= \left ( 6 + 1 \right ) \cdot e ^{ 9+ 3} =7e ^{ 12}$

Use implicit differentiation to find the slope of the tangent line to $\small y^3-lny=x^2$ at the point $\small (1,1)$ .

$\small 1$

$\small 0$

$\small -1$

$\small -2$

$\small 2$

Explanation

We must take the derivative $\small \left(\small \frac{\mathrm{d} y}{\mathrm{d} x}\right)$ because that will give us the slope. On the left side we'll get

$\small 3y^2\frac{\mathrm{d}y }{\mathrm{d} x}-(1/y)\frac{\mathrm{d} y}{\mathrm{d} x}$ , and on the right side we'll get $\small 2x$ .

We include the $\small \frac{\mathrm{d} y}{\mathrm{d} x}$ on the left side because $\small y$ is a function of $\small x$ , so its derivative is unknown (hence we are trying to solve for it!).

Now we can factor out a $\small \frac{\mathrm{d} y}{\mathrm{d} x}$ on the left side to get

$\small \small \frac{\mathrm{d} y}{\mathrm{d} x}(3y^2-1/y)=2x$ and divide by $\small 3y^2-1/y$ in order to solve for $\small \frac{\mathrm{d} y}{\mathrm{d} x}$ .

Doing this gives you

$\small \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2x}{3y^2-1/y}$ .

We want to find the slope at $\small (1,1)$ , so we can sub in $\small 1$ for $\small x$ and $\small y$ .

$\small \small \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2(1)}{3(1)^2-1/(1)}=2/2=1$ .

$g (x) = \frac{1}{4x - 7}$

Evaluate .

$- \frac{4}{9}$

$\frac{4}{9}$

$\frac{4}{3}$

$-\frac{4}{3}$

Undefined

Explanation

To find , substitute and use the chain rule:

$g (x) = \frac{1}{4x - 7}$

$g '(x) = \frac{\mathrm{d} }{\mathrm{d} x}\cdot \frac{1}{4x - 7}$

$= \frac{\mathrm{d} }{\mathrm{d} x}\cdot \frac{1}{u}$

$= \frac{\mathrm{d} u}{\mathrm{d} x}\cdot \frac{\mathrm{d} }{\mathrm{d} u} \frac{1}{u}$

$= \frac{\mathrm{d} }{\mathrm{d} x} (4x-7) \cdot \frac{\mathrm{d} }{\mathrm{d} u} u ^{-1}$

$=4 \cdot (-1 ) u ^{-1-1}$

$=-4 u ^{-2} = - \frac{4}{u^{2}}= - \frac{4}{(4x-7)^{2}}$

$g'(x)= - \frac{4}{(4x-7)^{2}}$

and

$g'(1)= - \frac{4}{(4 \cdot 1 -7)^{2}}$

$= - \frac{4}{(4 -7)^{2}}$

$= - \frac{4}{(-3)^{2}} = - \frac{4}{9}$

Screen shot 2016 03 31 at 11.11.35 am

Figure. Squircle of "radius" 1

A squircle is a curve in the xy plane that appears like a rounded square, but whose points satisfy the following equation (analogous to the Pythagorean theorem for a circle)

where the constant is the "radius" of the squircle.

Using implicit differentiation, obtain an expression for $\frac{dy}{dx}$ as a function of both and .

$-\frac{x^3}{y^3}$

$-\frac{4x^3}{y^3}$

$\frac{x^3}{y^3}$

$-\frac{x^4}{4y^3}$

$\frac{4x^3}{y^3}$

Explanation

Differentiate both sides of the equation with respect to , using the chain rule on the term:

$4x^3+4y^3\left (\frac{dy}{dx} \right )=0$

Then solve for $\frac{dy}{dx}$ as if it were our unknown:

$\frac{dy}{dx}=\frac{-4x^3}{4y^3}=-\frac{x^3}{y^3}$ .

Comparing this to the figure, our answer makes sense, because the slope of the squircle is wherever (as it crosses the axis) and undefined (vertical) wherever (as it crosses the axis). Lastly, we note that in the first quadrant (where and ), the slope of the squircle is negative, which is exactly what we observe in the figure.

A curve in the xy plane is given implictly by

Calculate the slope of the line tangent to the curve at the point .

$\frac{2}{5}$

$-\frac{16}{5}$

$\frac{14}{5}$

$-\frac{4}{5}$

$-\frac{3}{5}$

Explanation

Differentiate both sides with respect to using the chain rule and the product rule as:

$\left ((1)(y^2) + (x)(2y)\left (\frac{dy}{dx} \right ) \right )+\left ((2x)(y)+(x^2)\left(\frac{dy}{dx} \right ) \right )-1=0$

Then solve for as if it were our unknown:

$\left (\frac{dy}{dx} \right )(2xy+x^2)=1-2xy-y^2$

$\frac{dy}{dx}=\frac{1-2xy-y^2}{2xy+x^2}$ .

Finally, evaluate at the point to obtain the slope through that point:

$\frac{1-2(1)(-3)-(-3)^2}{2(1)(-3)+(1)^2} = \frac{-2}{-5}=\frac{2}{5}$ .

Find $\frac{\mathrm{d} \delta}{\mathrm{d} x}$ from the following equation:

$5=\delta e^{x^2}$ , where $\delta$ is a function of x.

$-2\delta x$

$2\delta x$

$-2\delta x e^{x^2}$

Explanation

To find the derivative of $\delta$ with respect to x, we must differentiate both sides of the equation with respect to x:

$0=2\delta xe^{x^2}+e^{x^2}\frac{\mathrm{d} \delta}{\mathrm{d} x}$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Solving for $\frac{\mathrm{d} \delta}{\mathrm{d} x}$ , we get

$\frac{\mathrm{d} \delta}{\mathrm{d} x}=\frac{-2\delta x e^{x^2}}{e^{x^2}}=-2\delta x$

Note that the chain rule was used because of the exponential and because $\delta$ is a function of x.

Find :

$f=\gamma \cos(x^2)$ , where $\gamma$ is a constant.

$-2x\gamma \sin(x^2)$

$2x\gamma \sin(x^2)$

$-2x\sin(x^2)$

$-\gamma \sin(x^2)$

$2x\gamma \cos(x^2)$

Explanation

The derivative of the function is equal to

$f'=-2x\gamma \sin(x^2)$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

The constant may seem intimidating, but we treat it as another constant!

Find the first derivative of the following function:

$f=\sec(x^2)+\ln(2x^2)$

$2x\sec(x^2)\tan(x^2)+\frac{2}{x}$

$\sec(x^2)\tan(x^2)+\frac{1}{2x^2}$

$2\sec(x^2)\tan(x^2)+\frac{2}{x}$

$2x\tan(x^2)+\frac{2}{x}$

Explanation

The derivative of the function is equal to

$f'=2x\sec(x^2)\tan(x^2)+\frac{4x}{2x^2}=2x\sec(x^2)\tan(x^2)+\frac{2}{x}$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sec(x)=\sec(x)\tan(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \ln(x)=\frac{1}{x}$

Note that the chain rule was used on the secant function as well as the natural logarithm function.

Find dx/dy by implicit differentiation:

$x^3+x\tan^{-1}(y)=e^y$

$\frac{e^y-\frac{x}{1+y^2}}{3x^2+\tan^{-1}(y)}$

$\frac{3x^2+\tan^{-1}(y)}{e^y-\frac{x}{1+y^2}}$

$\frac{\tan^{-1}(y)-\frac{x}{1+y^2}}{3x^2+e^y}$

$\frac{\frac{x}{1+y^2}+e^y}{5x^2-\tan^{-1}(y)}$

$\frac{3x^2+e^y}{\tan^{-1}(y)-\frac{x}{1+y^2}}$

Explanation

To find dx/dy we must take the derivative of the given function implicitly. Notice the term $x\tan^{-1}(y)$ will require the use of the Product Rule, because it is a composition of two separate functions multiplied by each other. Every other term in the given function can be derived in a straight-forward manner, but this term tends to mess with many students. Remember to use the Product Rule: