Ive been given this rule and asked to differentiate $\sqrt{x^2+1}$, however I am not sure what I am missing.It is said that if f is differentiable at x and f(x)>0.
$\frac{d}{\text{dx}}$$\sqrt{f(x)}=\frac{f'(x)}{2 \sqrt{f(x)}}$
What I thought would be correct is that:
$\frac{d}{\text{dx}}$$\sqrt{x^2+1}$ = $\frac{1}{2 \sqrt{x^2+1}}$,
However in the textbook the answer is:
$\frac{x}{ \sqrt{x^2+1}}$
I do not get how this square rule is made, and why $f'(x)$ is in the numerator.
Derive using the chain rule and the rule for polynomials: $$ f(x) =\sqrt{x^2+1} = (x^2+1)^{\frac{1}{2}},$$ then $$ f'(x) =\frac{1}{2}(x^2+1)^{-\frac{1}{2}} \cdot(2x) = x\cdot(x^2+1)^{-\frac{1}{2}} =\frac{x}{\sqrt{x^2+1}} $$
$$f'(x)=(\sqrt{x^2+1})'=\frac{1}{2\sqrt{x^2+1}}(x^2+1)'=\frac{2x}{2\sqrt{x^2+1}}=\frac{x}{\sqrt{x^2+1}}$$ Your first part is correct but you need to apply the chain rule, since you are differentiating $\sqrt{g(x)}$ not $\sqrt{x}$. There's a function under the square root instead of a variable.
set $$ w = x^2 + 1 \\ y = \sqrt{w} = w^\frac12 $$ so: $$ \frac{dy}{dw} = \frac12 w^{-\frac12} $$ and $$ \frac{dw}{dx} = 2x $$ now use the chain rule $$ \frac{dy}{dx} = \frac{dy}{dw} \times \frac{dw}{dx} $$
