Computation of Derivatives

Help Questions

AP Calculus BC › Computation of Derivatives

Questions 1 - 10

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

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:

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

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

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

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 $\small y$ .

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

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 $\small y$ .

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

Solve for $\frac{dy}{dx}$ if and .

$\frac{3}{2}$

$-\frac{3}{2}$

$\frac{2}{3}$

$-\frac{2}{3}$

None of the above

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 and , 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:

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

Solve for $\frac{dy}{dx}$ if and .

$\frac{3}{2}$

$-\frac{3}{2}$

$\frac{2}{3}$

$-\frac{2}{3}$

None of the above

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 and , 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:

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

$f (x) = 7x^{2} + 6x - 9$

Give .

$f'(x) = 14x^{2}+6x$

Explanation

$\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0, so

$f (x) = 7x^{2} + 6x - 9$

$f ' (x) = 7\cdot 2\cdot x^{2-1} + 6\cdot 1 + 0$

$f ' (x) = 14\cdot x^{1} + 6$

$f (x) = 7x^{2} + 6x - 9$

Give .

$f'(x) = 14x^{2}+6x$

Explanation

$\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0, so

$f (x) = 7x^{2} + 6x - 9$

$f ' (x) = 7\cdot 2\cdot x^{2-1} + 6\cdot 1 + 0$

$f ' (x) = 14\cdot x^{1} + 6$

$f(x)= 8x^{3}-7x^{2}+11$

Give .

Explanation

First, find the derivative of $f(x)= 8x^{3}-7x^{2}+11$ .

$\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0, so

$f(x)= 8x^{3}-7x^{2}+11$

$f'(x)= 3 \cdot 8x^{3-1}-2\cdot 7x^{2-1}+0$

$f'(x)= 24x^{2}-14x^{1}$

$f'(x)= 24x^{2}-14x$

Now, differentiate to get .

$f''(x)= 2 \cdot 24x^{2-1}-1\cdot 14x^{1-1}$

$f''(x)= 48x^{1}-14x^{0}$

$f(x)= 8x^{3}-7x^{2}+11$

Give .

Explanation

First, find the derivative of $f(x)= 8x^{3}-7x^{2}+11$ .

$\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0, so

$f(x)= 8x^{3}-7x^{2}+11$

$f'(x)= 3 \cdot 8x^{3-1}-2\cdot 7x^{2-1}+0$

$f'(x)= 24x^{2}-14x^{1}$

$f'(x)= 24x^{2}-14x$

Now, differentiate to get .

$f''(x)= 2 \cdot 24x^{2-1}-1\cdot 14x^{1-1}$

$f''(x)= 48x^{1}-14x^{0}$

Page 1 of 16

Return to subject

AI TutorYour personal study buddy

Question of the DayDaily practice to build your skills

FlashcardsStudy and memorize key concepts

WorksheetsBuild custom practice quizzes

AI SolverStep-by-step problem solutions

Learn by ConceptMaster concepts step by step

Diagnostic TestsAssess your knowledge level

Practice TestsTest your skills with exam questions

GamesLearn through free games