1. ## Calc chain rule HELP! Please!

Find derivative of:

2. Originally Posted by zaurm

Find derivative of: $\displaystyle (x^2+x+2)^{\sin x}$
Hint 1:

Rewrite it as $\displaystyle e^{\ln((x^2+x+2)^{\sin x})} = e^{\sin x(\ln(x^2+x+2))}$.

Now use the chain rule.

Hint 2:
Spoiler:

The derivative is:

$\displaystyle e^{\sin x(\ln(x^2+x+2))}*\frac{d}{dx}[\sin x(\ln(x^2+x+2))]$

$\displaystyle \frac{d}{dx}[\sin x(\ln(x^2+x+2))] = \sin x*\frac{d}{dx}[\ln(x^2+x+2)] + \ln(x^2+x+2)*\frac{d}{dx}[\sin x]$

Solution:
Spoiler:

The derivative is:

$\displaystyle e^{\sin x(\ln(x^2+x+2))}*\frac{d}{dx}[\sin x(\ln(x^2+x+2))]$

$\displaystyle \frac{d}{dx}[\sin x(\ln(x^2+x+2))] = \sin x*\frac{d}{dx}[\ln(x^2+x+2)] + \ln(x^2+x+2)*\frac{d}{dx}[\sin x]$

$\displaystyle = \sin x* \frac{2x+1}{x^2+x+2} + \ln(x^2+x+2)*\cos x = \frac{(2x+1)\sin x}{x^2+x+2} + \ln(x^2+x+2)\cos x$

So the whole derivative is (bearing in mind that $\displaystyle e^{\sin x(\ln(x^2+x+2))} = (x^2+x+2)^{\sin x}$):

$\displaystyle (x^2+x+2)^{\sin x}\left(\frac{(2x+1)\sin x}{x^2+x+2} + \ln(x^2+x+2)\cos x\right)$