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AP Calculus BC : AP Calculus BC

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Example Questions

Example Question #21 : Chain Rule And Implicit Differentiation

Find :

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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!

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Example Question #61 : Computation Of Derivatives

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

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

Possible Answers:

$2\delta x$

$-2\delta x$

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

-2 x

Correct answer:

$-2\delta x$

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.

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Example Question #21 : Chain Rule And Implicit Differentiation

Find the first derivative of the following function:

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

Possible Answers:

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

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

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

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

Correct answer:

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

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.

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Example Question #62 : Computation Of Derivatives

Find $\frac{\mathrm{d} y}{\mathrm{d} x}$ :

xe^y=y^2

Possible Answers:

$\frac{e^y}{2y-2xe^y}$

$\frac{y}{x}$

$\frac{-xe^{2y}}{y}$

$\frac{e^y}{-2y+2xe^y}$

$\frac{e^y}{2y+2xe^y}$

Correct answer:

$\frac{e^y}{2y-2xe^y}$

Explanation:

To determine $\frac{\mathrm{d} y}{\mathrm{d} x}$ , we must take the derivative of both sides of the equation with respect to x:

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

The derivatives were found using the following rules:

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

Rearranging and solving for $\frac{\mathrm{d} y}{\mathrm{d} x}$ , we get

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

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Example Question #22 : Chain Rule And Implicit Differentiation

A curve in the xy plane is given implictly by

xy^2+x^2y-x=5 .

Calculate the slope of the line tangent to the curve at the point (1, -3) .

Possible Answers:

$-\frac{16}{5}$

$\frac{2}{5}$

$\frac{14}{5}$

$-\frac{4}{5}$

$-\frac{3}{5}$

Correct answer:

$\frac{2}{5}$

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 dy/dx 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 dy/dx at the point (1,-3) to obtain the slope through that point:

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

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Example Question #161 : Derivatives

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)

x^4+y^4=R^4

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 .

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Differentiate both sides of the equation with respect to , using the chain rule on the y^4 term:

x^4+y^4=R^4

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

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 x=0 (as it crosses the axis) and undefined (vertical) wherever y=0 (as it crosses the axis). Lastly, we note that in the first quadrant (where x>0 and y>0 ), the slope of the squircle is negative, which is exactly what we observe in the figure.

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Example Question #1 : Derivatives Of Parametrics

Find the derivative of the following set of parametric equations:

x=2t^3+5t^2-11t+3

y=7t^2-4t-18

Possible Answers:

$\frac{t^2-7t+8}{6t-2}$

$\frac{5t^2+6t+3}{2t-11}$

$\frac{2(7t-2)}{6t^2+10t-11}$

$\frac{-14t+1}{8t^2+5t-2}$

$\frac{4(3t+4)}{2t^2+4t-3}$

Correct answer:

$\frac{2(7t-2)}{6t^2+10t-11}$

Explanation:

We start by taking the derivative of x and y with respect to t, as both of the equations are only in terms of this variable:

$\frac{dx}{dt}=6t^2+10t-11$

$\frac{dy}{dt}=14t-4$

The problem asks us to find the derivative of the parametric equations, dy/dx, and we can see from the work below that the dt term is cancelled when we divide dy/dt by dx/dt, leaving us with dy/dx:

$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{dy}{dt}\frac{dt}{dx}=\frac{dy}{dx}$

So now that we know dx/dt and dy/dt, all we must do to find the derivative of our parametric equations is divide dy/dt by dx/dt:

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{14t-4}{6t^2+10t-11}=\frac{2(7t-2)}{6t^2+10t-11}$

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Example Question #2 : Derivatives Of Parametrics

Solve for $\frac{dy}{dx}$ if x=2t-5 and y=3t+8 .

Possible Answers:

$\frac{2}{3}$

$-\frac{3}{2}$

None of the above

$\frac{3}{2}$

$-\frac{2}{3}$

Correct answer:

$\frac{3}{2}$

Explanation:

We can determine that $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ since the terms will cancel out in the division process.

Since x=2t-5 and y=3t+8 , we can use the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ to derive

$\frac{dx}{dt}=(1)2t^{1-1}-(0)5=2$ and $\frac{dy}{dt}=(1)3t^{1-1}+(0)8=3$ .

Thus:

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{3}{2}}$ .

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Example Question #1 : Derivatives Of Parametric, Polar, And Vector Functions

Find the derivative of the following polar equation:

$r=3+8sin\theta$

Possible Answers:

$\frac{8cos\theta sin\theta+3sin\theta }{4sin^2\theta-3sin\theta-8cos^2\theta}$

$\frac{16sin^2\theta+3cos\theta }{8cos^2\theta-3cos\theta-8sin^2\theta}$

$\frac{16cos\theta sin\theta+3cos\theta }{8cos^2\theta-3sin\theta-8sin^2\theta}$

$\frac{8cos\theta sin\theta-6cos\theta }{4cos^2\theta+6sin\theta+3sin^2\theta}$

$\frac{8cos\theta sin^2\theta+3cos^2\theta }{4cos^2\theta+3sin\theta-6sin^2\theta}$

Correct answer:

$\frac{16cos\theta sin\theta+3cos\theta }{8cos^2\theta-3sin\theta-8sin^2\theta}$

Explanation:

Our first step in finding the derivative dy/dx of the polar equation is to find the derivative of r with respect to $\theta$ . This gives us:

$\frac{dr}{d\theta}=8cos\theta$

Now that we know dr/d $\theta$ , we can plug this value into the equation for the derivative of an expression in polar form:

$\frac{dy}{dx}=\frac{\frac{dr}{d\theta}sin\theta+rcos\theta }{\frac{dr}{d\theta }cos\theta-rsin\theta }=\frac{8cos\theta sin\theta+(3+8sin\theta )cos\theta}{8cos^2\theta-(3+8sin\theta)sin\theta }$

Simplifying the equation, we get our final answer for the derivative of r:

$\frac{dy}{dx}=\frac{16cos\theta sin\theta +3cos\theta }{8cos^2\theta -3sin\theta -8sin^2\theta }$

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Example Question #2 : Derivatives Of Parametric, Polar, And Vector Functions

Find the derivative $\frac{dy}{dx}$ of the polar function $r(\theta)=3-4sin(\theta)$ .

Possible Answers:

$\frac{dy}{dx}=\frac{-4cos(\theta)sin(\theta)+(3-4sin(\theta))cos(\theta)}{-4cos^2(\theta)-(3-4sin(\theta))sin(\theta)}$

$\frac{dy}{dx}=\frac{-4cos(\theta)cos(\theta)+(3-4sin(\theta))sin(\theta)}{-4cos(\theta)sin(\theta)-(3-4sin(\theta))cos(\theta)}$

$\frac{dy}{dx}=4cos(\theta)$

$\frac{dy}{dx}=-4cos(\theta)$

Correct answer:

$\frac{dy}{dx}=\frac{-4cos(\theta)sin(\theta)+(3-4sin(\theta))cos(\theta)}{-4cos^2(\theta)-(3-4sin(\theta))sin(\theta)}$

Explanation:

The derivative of a polar function is found using the formula

$\frac{dy}{dx}=\frac{r'(\theta)sin(\theta)+r(\theta)cos(\theta)}{r'(\theta)cos(\theta)-r(\theta)sin(\theta)}$

The only unknown piece is $r'(\theta)$ . Recall that the derivative of a constant is zero, and that

$\frac{d}{d\theta}sin(\theta)=cos(\theta)$ , so

$r'(\theta)=-4cos(\theta)$

Substiting $r'(\theta)\ and\ r(\theta)$ this into the derivative formula, we find

$\frac{dy}{dx}=\frac{-4cos(\theta)sin(\theta)+(3-4sin(\theta))cos(\theta)}{-4cos(\theta)cos(\theta)-(3-4sin(\theta))sin(\theta)}$

$\frac{dy}{dx}=\frac{-4cos(\theta)sin(\theta)+(3-4sin(\theta))cos(\theta)}{-4cos^2(\theta)-(3-4sin(\theta))sin(\theta)}$

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AP Calculus BC : AP Calculus BC

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #21 : Chain Rule And Implicit Differentiation

Example Question #61 : Computation Of Derivatives

Example Question #21 : Chain Rule And Implicit Differentiation

Example Question #62 : Computation Of Derivatives

Example Question #22 : Chain Rule And Implicit Differentiation

Example Question #161 : Derivatives

Example Question #1 : Derivatives Of Parametrics

Example Question #2 : Derivatives Of Parametrics

Example Question #1 : Derivatives Of Parametric, Polar, And Vector Functions

Example Question #2 : Derivatives Of Parametric, Polar, And Vector Functions

All AP Calculus BC Resources

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