Understanding Derivatives of Sums, Quotients, and Products - High School Math

Card 0 of 5

Question

Find the derivative of the following function:

$f(x) = x^3 + \frac{1}{x}$

Answer

Since this function is a polynomial, we take the derivative of each term separately.

From the power rule, the derivative of

is simply

We can rewrite $\frac{1}{x}$ as

$x^{-1}$

and using the power rule again, we get a derivative of

$-x^{-2}$ or $-\frac{1}{x^2}$

So the answer is

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

Compare your answer with the correct one above

Question

Which of the following best represents $\frac{\mathrm{d} }{\mathrm{d} x}\frac{f(x)}{g(x)}$ ?

Answer

The question is just asking for the Quotient Rule formula.

Recall the Quotient Rule is the bottom function times the derivative of the top minus the top function times the derivative of the bottom all divided by the bottom function squared.

Given,

$\frac{\mathrm{d} }{\mathrm{d} x}\frac{f(x)}{g(x)}$

the bottom function is and the top function is . This makes the bottom derivative and the top derivative .

Substituting these into the Quotient Rule formula resulting in the following.

$\frac{\mathrm{d} }{\mathrm{d} x}\frac{f(x)}{g(x)}=\frac{g(x)\cdot f'(x)-f(x)\cdot g'(x)}{(g(x))^2}$

Compare your answer with the correct one above

Question

What is $\frac{\mathrm{d} }{\mathrm{d} x}f(x)*g(x)\mathrm{d x} ?$

Answer

The chain rule is "first times the derivative of the second plus second times derivative of the first".

In this case, that means $\frac{\mathrm{d} }{\mathrm{d} x}f(x)*g(x)\mathrm{d x} =f(x)g'(x)+f'(x)g(x)$ .

Compare your answer with the correct one above

Question

What is the first derivative of $5x^4+\sin x$ ?

Answer

Since we're adding terms, we take the derivative of each part separately. For , we can use the power rule, which states that we multiply the variable by the current exponent and then lower the exponent by one. For sine, we use our trigonometric derivative rules.

Remember, $\frac{\mathrm{d} }{\mathrm{d} x} \sin x=\cos x$ .

$\frac{\mathrm{d} }{\mathrm{d} x}5x^4+\sin x=(4*5x^{4-1})+\cos x$

$\frac{\mathrm{d} }{\mathrm{d} x}5x^4+\sin x=(20x^{3})+\cos x$

Compare your answer with the correct one above

Question

What is the second derivative of $5x^4+\sin x$ ?

Answer

To find the second derivative, we need to start by finding the first one.

Since we're adding terms, we take the derivative of each part separately. For , we can use the power rule, which states that we multiply the variable by the current exponent and then lower the exponent by one. For sine, we use our trigonometric derivative rules.

Remember, $\frac{\mathrm{d} }{\mathrm{d} x} \sin x=\cos x$ .

$\frac{\mathrm{d} }{\mathrm{d} x}5x^4+\sin x=(45x^{4-1})+\cos x$

$\frac{\mathrm{d} }{\mathrm{d} x}5x^4+\sin x=(20x^{3})+\cos x$

Now we repeat the process, but using $20x^3+\cos x$ as our equation.

$\frac{\mathrm{d} }{\mathrm{d} x} \cos x=-\sin x$

$\frac{\mathrm{d} }{\mathrm{d} x}20x^3+\cos x=(320x^{3-1})-\sin x$

$\frac{\mathrm{d} }{\mathrm{d} x}20x^3+\cos x=(60x^{2})-\sin x$

Compare your answer with the correct one above

Tap the card to reveal the answer