Differentiate $\sin \sqrt{x^2+1} $with respect to $x$?

Differentiate $$ \sin \sqrt{x^2+1} $$ with respect to $x$?

Can someone please help me with question, im very lost.

$$ \dfrac {d}{dx}\sin \sqrt{x^2+1} $$ since $\dfrac {d}{dx}\sin x=\cos x,\dfrac {d}{dx} x^n=n\cdot x^{n-1},\dfrac{d}{dx}C=0$ C is Constant

so $$ \dfrac {d}{dx}\sin \sqrt{x^2+1}\implies\cos \sqrt{x^2+1}\dfrac {d}{dx}\sqrt{x^2+1}$$ like this continue the differentiation of functions (here differentiate $\sqrt{x^2+1}$). Just try to solve it is very simple then.

Ayush is incorrect, the chain rule is actually:

$$\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$$

Which would make your answer

$$\frac{d}{dx} \sin\sqrt{x^2+1} = \cos\sqrt{x^2+1} \cdot \frac{d}{dx}\sqrt{x^2+1} = \cos\sqrt{x^2+1} \cdot \frac{x}{\sqrt{x^2+1}} $$.

You can check the answer with wolfram alpha.

OK, since you're not that familiar with the chain rule, let me simplify that for you.

Chain rule of differentiation: $$\frac{d}{dx}f\left(g\left(x \right) \right) =\left( \left.\frac{d}{dy}f(y)\right|_{y=f(x)} \right) \cdot \left( \frac{d}{dx}g(x) \right)$$

So,

$$ \frac{d}{dx}\left(\sin \sqrt{x^2+1} \right)= \frac{d}{dy}\left.\left(\sin y \right)\right|_{y=\sqrt{x^2+1}} \cdot\frac{d}{dx}\left( \sqrt{x^2+1} \right) $$

Your answer: $$=\cos (\sqrt{x^2+1})\cdot\frac{x}{\sqrt {x^2 + 1}}$$