Chapters

More Practice

In the previous section, we learned useful formulae for solving derivatives. They are called the

Product Rule: $\displaystyle{\frac{df(x)g(x)}{dx} = f'(x)g(x)+g'(x)f(x)}$

and the

Quotient Rule: $\displaystyle{\frac{d \frac{f(x)}{g(x)}}{dx} = \frac{f'(x)g(x)-g'(x)f(x)}{(g(x))^2}}$

Now it's time to put these formulas to use.

Here are some practice problems to use these formulas.

$\displaystyle{\frac{d \tan{x}}{dx} = }$

Hint 1

What's another way to write $\tan x$

Hint 2

Which rule should we use, product or quotient?

Solution 1

Use the division rule to solve the derivative:

\tan x = \frac{\sin x}{\cos x}

f(x) \rightarrow \sin x

f'(x) \rightarrow \cos x

g(x) \rightarrow \cos x

g'(x) \rightarrow - \sin x

\frac{\cos x\cdot\cos x+\sin x\cdot\sin x}{\cos^2x}=\frac{\cos^2x+\sin^2x}{\cos^2x}

Use the pythagorean identity:

=\frac{1}{\cos^2 x} = \sec^2x

$\displaystyle{\frac{d f(x)e^x}{dx}}= e^x \cdot$

Hint 1

Which rule should we use? Quotient, or product?

Solution 1

Use the product rule to solve the derivative:

e^x \rightarrow g(x)

f(x)\rightarrow f(x)

f\left(x\right)e^x+f'\left(x\right)e^x

Distribute:

=\boxed{e^x\left[f\left(x\right)+f'\left(x\right)\right]}

$\displaystyle{\frac{d \sec x}{dx}}=$

Hint 1

\sec x = \frac{1}{\cos x}

Solution 1

Use the Quotient Rule to solve the derivative:

\sec x = \frac{1}{\cos x}

f(x) \rightarrow 1

f'(x) \rightarrow 0

g(x) \rightarrow \cos x

g'(x) \rightarrow - \sin x

\frac{\left(0\right)\left(\cos x\right)-\left(-\sin x\right)\left(1\right)}{\cos^2x}=\frac{\sin x}{\cos^2x}=\boxed{\tan x\sec x}

$\displaystyle{\frac{d\csc x}{dx}}=$

Hint 1

\csc x = \frac{1}{\sin x}

Solution 1

Use the Quotient Rule to solve the derivative:

\sec x = \frac{1}{\sin x}

f(x) \rightarrow 1

f'(x) \rightarrow 0

g(x) \rightarrow \sin x

g'(x) \rightarrow \cos x

\frac{\left(0\right)\left(\sin x\right)-\left(\cos x\right)\left(1\right)}{\sin^2x}=-\frac{\cos x}{\sin^2x}=\boxed{-\csc x\cot x}