$\frac{d}{dx} \sqrt{x+2}$

Question

Forgive me for my simple question, calculus from Engineering school is about ten years in the past for me.

$f(x) = \sqrt{x+2}$

What is $f'(x)$?

If it were just $f(x) = \sqrt{x}$ it would be easy, because $\sqrt{x} = x^\frac{1}{2}$, but the case of $f(x) = \sqrt{x+2}$ is different because of $+2$, correct? Or am I wrong? :/

Answer 1

Use the chain rule$$\frac {dy}{dx}=\frac {dy}{du}\cdot\frac {du}{dx}$$Or, in a more recognizable form with $f(x)$ and $g(x)$$$f(g(x))'=g'(x)f'(g(x))$$Where $g(x)=x+2$ and $f(x)=\sqrt x$.

Answer 2

Use chain rule. $$\frac {d}{dx} f(g(x)) = f'(g(x))g'(x)$$ This leads to $$\frac{d}{dx} \sqrt{x+2} = \frac {1}{2 \sqrt{x+2}} \frac{d}{dx} (x+2) = \frac {1}{2 \sqrt{x+2}} $$

Answer 3

wrtie $$f(x)=(x+2)^{1/2}$$ then $$f'(x)=\frac{1}{2}(x+2)^{1/2-1}$$

Answer 4

$$f(x)=(x+2)^{\frac{1}{2}},$$ then by the chain rule

$$f'(x)=\frac{1}{2}(x+2)^{-\frac{1}{2}}.$$

Answer 5

We have $$f(x)=\sqrt{x+2}=(x+2)^{1/2}$$ and since the derivative of $x+2$ is $1$, by the Chain Rule, $$f'(x)=\frac12(x+2)^{-1/2}\cdot1=\boxed{\frac1{2\sqrt{x+2}}}$$