Question

question number 10. (4.00 points)
differentiate the following: $f(x) = \tan^{-1}(x + 4)$
$\circ f(x) = \frac{1}{x^2 - 8\\,x + 17}$
$\circ f(x) = \frac{1}{x^2 + 8\\,x + 17}$
$\circ f(x) = \frac{1}{2\\,x^2 + 8\\,x + 16}$
$\circ f(x) = \frac{1}{2\\,x^2 + 8\\,x + 15}$
$\circ f(x) = \frac{1}{x^2 + 4\\,x + 17}$
$\circ$none of the above.
question number 11. (4.00 points)
we know $f(x)$ is a differentiable function. we are given its first derivativ

Explanation:

Step1: Recall the derivative of arctangent

The derivative of \( \tan^{-1}(u) \) with respect to \( x \) is \( \frac{1}{1 + u^2} \cdot u' \), where \( u \) is a function of \( x \).
Here, \( u = x + 4 \), so \( u' = 1 \).

Step2: Substitute \( u \) into the formula

Using the chain rule, \( f'(x) = \frac{1}{1 + (x + 4)^2} \cdot 1 \).

Step3: Expand the denominator

Expand \( (x + 4)^2 \): \( (x + 4)^2 = x^2 + 8x + 16 \).
Then, \( 1 + (x + 4)^2 = 1 + x^2 + 8x + 16 = x^2 + 8x + 17 \).

Step4: Write the derivative

So, \( f'(x) = \frac{1}{x^2 + 8x + 17} \).

Answer:

\( f'(x) = \frac{1}{x^2 + 8x + 17} \) (corresponding to the option \( f'(x) = \frac{1}{x^2 + 8x + 17} \))