Question

I’m in AP Calc AB and can’t figure this out. Any idea?

Answer 1

Answer::

`f^(\prime)(x)=7x\sec x\tan x+7\sec x+(1)/(x)`

Explanation:

Given f(x) defined below:

`f(x)=\ln x+7x\sec x`

The derivative is calculated below.

`\begin{gathered} (d)/(dx)\lbrack f(x)\rbrack=(d)/(dx)\lbrack\ln x+7x\sec x\rbrack \\ =(d)/(dx)\lbrack\ln x\rbrack+(d)/(dx)\lbrack7x\sec x\rbrack \\ Take\text{ the constant 7 outside the derivative sign.} \\ =$$\textcolor{red}{(d)/(dx)\lbrack\ln x\rbrack}$$+7(d)/(dx)\lbrack x\sec x\rbrack \\ \text{The derivative of }\ln (x)=(1)/(x),\text{ therefore:} \\ $$\textcolor{red}{(d)/(dx)\lbrack\ln x\rbrack}$$+7(d)/(dx)\lbrack x\sec x\rbrack=$$\textcolor{red}{(1)/(x)}$$+7(d)/(dx)\lbrack x\sec x\rbrack\cdots(1) \end{gathered}`

Next, we find the derivative of x sec x using the product rule.

`\begin{gathered} (d)/(dx)\lbrack x\sec x\rbrack=x$$\textcolor{blue}{(d)/(dx)\lbrack\sec x\rbrack}$$+\sec x(d)/(dx)\lbrack x\rbrack\text{ } \\ The\text{ derivative of sec(x), }\text{\textcolor{red}{ }}\textcolor{red}{(d)/(dx)\lbrack\sec x\rbrack=\sec x\tan x} \\ =x$$\textcolor{blue}{\lbrack\sec x\tan x\rbrack}$$+\sec x \end{gathered}`

Substitute the result into equation (1) above.

`\begin{gathered} (1)/(x)+7(d)/(dx)\lbrack x\sec x\rbrack=(1)/(x)+7(x\sec x\tan x+\sec x) \\ =7x\sec x\tan x+7\sec x+(1)/(x) \end{gathered}`

Therefore:

`f^(\prime)(x)=7x\sec x\tan x+7\sec x+(1)/(x)`