### Calculus

Topics within Calculus

## 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.

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

What's another way to write $\tan x$
Which rule should we use, product or quotient?

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:

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

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

Which rule should we use? Quotient, or product?

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:

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

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

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

Use the Quotient Rule to solve the derivative:

$$\sec x = \frac{1}{\cos x}$$
$$f(x) \rightarrow 1$$ $$f'(x) \rightarrow 0$$
$$g(x) \rightarrow \cos x$$ $$g'(x) \rightarrow - \sin x$$

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

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

$$\csc x = \frac{1}{\sin x} Use the Quotient Rule to solve the derivative:$$\sec x = \frac{1}{\sin x}f(x) \rightarrow 1f'(x) \rightarrow 0g(x) \rightarrow \sin xg'(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}