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Calculus 1 : Equations

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Example Questions

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Example Question #381 : Equations

Find the derivative of f(x)=(2x^2-1)(4x+1) .

Possible Answers:

$f{}'(x)=24x^2+4x-\frac{1}{4}$

$f{}'(x)=24x^2-4$

$f{}'(x)=4x^2+4x-4$

$f{}'(x)=24x^2+4x$

$f{}'(x)=24x^2+4x-4$

Correct answer:

$f{}'(x)=24x^2+4x-4$

Explanation:

To find the derivative of this function, recall the product rule. You multiply the first expression by the derivative of the second and the add that to the product of derivative of the first expression by the second. Therefore, $f{}'(x)=(2x^2-1)(4)+(4x+1)(4x)$ . Then, multiply and combine like terms so that your final answer is: $f{}'(x)=24x^2+4x-4$ .

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Example Question #382 : Equations

$f(x)= 4x^{8}+3x^{4}-2x^{3}+12$

Find the derivative.

Possible Answers:

$f'(x)=32x^{7}+12x^{3}-6x^{2}+12$

$f'(x)=32x^{8}+12x^{4}-6x^{3}+12$

$32x^{7}+12x^{3}-6x^{2}$

$f'(x)= 12x^{7}+7x^{3}-5x^{2}$

Correct answer:

$32x^{7}+12x^{3}-6x^{2}$

Explanation:

$f'(x)=32x^{7}+12x^{3}-6x^{2}$ is the correct answer.

When taking the derivative of a function, multiply the exponent by the constant in front of and subtract one from the exponent.

So, $4x^{8}$ becomes $( 8\times4\left\right)x^{(8-1)}=32x^{7}$ and so on maintaining the positive and negative symbols.

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Example Question #383 : Equations

$f(x)= (12x^{2}-4x)^{5}$

Find the derivative.

Possible Answers:

$f'(x)=5(24x-4)(12x^{2}-4x)$

$f'(x)=5(12x^{2}-4x)^{4}$

$f'(x)=5(24x-4)^{4}$

$f'(x)=(24x-4)(12x^{2}-4x)$

$f'(x)=5(24x-4)(12x^{2}-4x)^{4}$

Correct answer:

$f'(x)=5(24x-4)(12x^{2}-4x)^{4}$

Explanation:

Here, you are required to use the chainrule because there is a function inside of a function.

Therefore, you first differentiate $f(x)= (12x^{2}-4x)^{5}$ as a whole treating the inside of the parenthesis as just .

You get $5(12x^{2}-4x)^{4}$ .

Then, you apply the chain rule and differentiate the function inside the parenthesis:

You get 24x-4 .

So your full answer is: $f'(x)=5(24x-4)(12x^{2}-4x)^{4}$

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Example Question #384 : Equations

$f(x)= 4x^{3}-12x$

What is the slope of this curve at x=0 ?

Possible Answers:

Correct answer:

Explanation:

To find the slope of a curve at a certain point, you must first find the derivative of that curve.

The derivative of this curve is $f'(x)=12x^{2}-12$ .

Then, plug in for into the derivative and you will get as the slope of f(x) at x=0 .

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Example Question #381 : Equations

Find the general solution to the following differential equation:
$\frac{\mathrm{dy} }{\mathrm{d} x}=x^3y$

Possible Answers:

$y=\ln(\frac{x^4}{4}))+C$

$y=e^{\frac{x^4}{4}}+C$

$y=Ce^{\frac{x^4}{4}}$

$y=\frac{x^4}{4}\cdot \frac{y^2}{2}+C$

Correct answer:

$y=Ce^{\frac{x^4}{4}}$

Explanation:

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Example Question #132 : Solutions To Differential Equations

f(x)=g(h(x))

$g(x)=x^{3}+x^{2}-1$

$h(x)=(x^{2}-2)$

Find f'(x) .

Possible Answers:

f'(x)=6x(x^2-2)^2+4x(x^2-2)

$f'(x)=(x^{3}+x^{2}-1)(3x+2)(2x)$

$f'(x)=2(x^{3}+x^{2}-1)$

$f'(x)= (x^{2}-2)(3x^{2}-4)$

Correct answer:

f'(x)=6x(x^2-2)^2+4x(x^2-2)

Explanation:

To find f'(x) first find f(x)=g(h(x)) .

Since,

$g(x)=x^{3}+x^{2}-1$

$h(x)=(x^{2}-2)$

then,

f(x)=(x^2-2)^3+(x^2-2)^2-1 .

We have to use the chain rule and power rule to solve this problem:

Chain Rule: $f'(x)=g'(h(x))\cdot(h'(x))$

Power Rule: $(x^n)'=nx^{n-1}$

Therefore,

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

f'(x)=6x(x^2-2)^2+4x(x^2-2)

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Example Question #1431 : Functions

Solve for f''(t) at t=4 .

${f}(t)= 2t^{3}-3t$

Possible Answers:

f''(4)=96

f{}''(4)=48

f{}''(4)=116

f{}''(4)= 93

Correct answer:

f{}''(4)=48

Explanation:

To solve this problem, we must first find f'(t) by deriving the original equation.

In this case, we just need to use the power rule which states,

[f(x)g(x)]'=f(x)g'(x)+g(x)f'(x) .

${f}(t)= 2t^{3}-3t$

$f{}'(t)= 6t^{2}-3$

We can then derive f'(t), again using just the power rule which states $(x^n)'=nx^{n-1}$ .

f''(t)= 12t

When we plug 4 into this final function, we arrive at the answer:

f''(4)= 12(4)=48

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Example Question #2461 : Calculus

Finding where the first derivative is positive tells us what about an equation?

Possible Answers:

Where the function is decreasing.

Where the function is concave up.

Where the function is increasing.

Where the function is concave down.

Correct answer:

Where the function is increasing.

Explanation:

Taking the first derivative gives us the function for the slope for our entire equation. Thus, finding where the slope is positive tells us where the function is increasing.

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Example Question #135 : Solutions To Differential Equations

Find the general solution to the following differential equation:

$\frac{\mathrm{dy} }{\mathrm{d} x}=x^2y+x^3y$

Possible Answers:

$y=e^{\frac{x^3}{3}+\frac{x^4}{4}}$

$y=Ce^{\frac{x^3}{3}+\frac{x^4}{4}}$

$y=Ce^{{x^3}+{x^4}$

$y=Ce^{{x^2}+x^3}$

Correct answer:

$y=Ce^{\frac{x^3}{3}+\frac{x^4}{4}}$

Explanation:

To find the general solution for the separable differential equation, we must move x and dx, y and dy to the same sides:

$\frac{dy}{y}=(x^3+x^2)dx$

Next, we integrate both sides:

$\ln\left | y\right |=\frac{x^3}{3}+\frac{x^4}{4}+C$

The integrals were found using the following rules:

$\int \frac{dx}{x}=\ln\left | x\right |+C$ , $\int x^n dx=\frac{x^{n+1}}{n+1}+C$

To solve for y, we must exponentiate both sides of the equation:

$y=e^{\frac{x^3}{3}+\frac{x^4}{4}+C}=e^{\frac{x^3}{3}+\frac{x^4}{4}}\cdot e^C=Ce^{\frac{x^3}{3}+\frac{x^4}{4}}$

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Example Question #136 : Solutions To Differential Equations

Find the general solution for the following differential equation:

$\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{x^3}{e^y}$

Possible Answers:

$y=C\ln(\frac{x^4}{4})$

$y=\ln(\frac{x^4}{4})+C$

$y=e^{x^4}+C$

$y=(\frac{x^4}{4}\cdot e^y) +C$

Correct answer:

$y=C\ln(\frac{x^4}{4})$

Explanation:

To solve the separable differential equation, we must move x and dx, y and dy to their respective sides, and integrate both sides:

e^y dy=x^3 dx

$\int e^y dy=e^y+C$

$\int x^3 dx= \frac{x^4}{4}+C$

The integrations were performed using the following rules:

$\int e^x dx=e^x+C$

$\int x^ndx=\frac{x^{n+1}}{n+1}+C$

The equation then becomes

$y=\frac{x^4}{4}+C$

Note that we combine all constants of integration to a single C.

Now, we solve for y by taking the natural logarithm of both sides of the equation:

$y=\ln(\frac{x^4}{4}+C)$

Simplifying, we get

$y=C\ln(\frac{x^4}{4})$ .

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Calculus 1 : Equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #381 : Equations

Example Question #382 : Equations

Example Question #383 : Equations

Example Question #384 : Equations

Example Question #381 : Equations

Example Question #132 : Solutions To Differential Equations

Example Question #1431 : Functions

Example Question #2461 : Calculus

Example Question #135 : Solutions To Differential Equations

Example Question #136 : Solutions To Differential Equations

All Calculus 1 Resources

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