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

   Hint 1

   What's another way to write $\tan x$
   Hint 2

   Which rule should we use, product or quotient?
   Show/Hide Solution

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

   Hint

   Which rule should we use? Quotient, or product?
   Show/Hide Solution

   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}}= $

   Hint

   $$\sec x = \frac{1}{\cos x}$$
   Show/Hide Solution

   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}}= $

   Hint

   $$\csc x = \frac{1}{\sin x}
   Show/Hide Solution

   Use the Quotient Rule to solve the derivative:

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

   $$f(x) \rightarrow 1$$ $$f'(x) \rightarrow 0$$

   $$g(x) \rightarrow \sin x$$ $$g'(x) \rightarrow \cos x$$

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