Question

If f(x)= $\sqrt{x^2-4}$ and g(x)=3x-2, then the derivative of f(g(x)) at x=3 is

Answer 1

Answer:

$\displaystyle \frac{d}{dx}\Big|_{x=3} & = \frac{9(3)-6}{\sqrt{(3(3)-2)^2-4}} = \frac{7\sqrt{5}}{5}$

Step-by-step explanation:

We are given the two functions:

$\displaystyle f(x) = \sqrt{x^2 -4} \text{ and } g(x) = 3x-2$

And we want to find:

$\displaystyle \frac{d}{dx}\left[f(g(x))\right] \text{ at } x = 3$

Recall that from the chain rule:

$\displaystyle \frac{d}{dx}\left[ f(g(x))\right] = f'(g(x)) \cdot g'(x)$

Hence, find both f'(x) and g'(x).

The derivative of f(x) can be determined using the chain rule:

$\displaystyle \begin{aligned} f'(x) &= \frac{d}{dx}\left[ \sqrt{x^2-4}\right]\\ \\ & = \frac{1}{2\sqrt{(x^2-4)}}\cdot \frac{d}{dx}\left[x^2-4\right] \\ \\ & = \frac{x}{\sqrt{x^2-4}} \end{aligned}$

Because g(x) is simply a linear function, its derivative will be the slope of the linear function at any point. That is:

$\displaystyle g'(x) = 3$

Hence:

$\displaystyle \begin{aligned} \frac{d}{dx}\left[f(g(x))\right] & = \frac{(g(x))}{\sqrt{(g(x))^2-4}} \cdot (3) \\ \\ & = \frac{9x-6}{\sqrt{(3x-2)^2-4}} \end{aligned}$

Then at the point x = 3, the derivative is:

$\displaystyle \frac{d}{dx}\Big|_{x=3} & = \frac{9(3)-6}{\sqrt{(3(3)-2)^2-4}} = \frac{7\sqrt{5}}{5}$