Question: We all know this rule:

> 
> $\text{If } y = a^{f(x)} \text{ then } y' = a^{f(x)} \: f'(x)\ln a$
> 
> 
> 

In my book there is the example:

Find $\frac{d}{dx}\left((x^{2} + 1)^{\sin x}\right)$

According to the rule, my answer is:

$(x^{2} + 1)^{\sin x} \cdot \cos x \cdot \ln(x^{2} + 1)$

But the answer of the book was:

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

So, where did $\frac{2x \sin x}{x^{2} + 1}$ come from?

Answer: 
I will assume that this is not homework and go ahead to give a full solution.

As others have said, your rule only works for constant $a$. Following the hint in danielson's answer and setting $y=f(x)$, you have $$\log y=\sin x\log\left(1+x^2\right)$$
and differentiating both sides with respect to $x$ and using the chain and product rules gives $$\frac1y\frac{dy}{dx}=\cos x\log\left(1+x^2\right)+\frac{2x\sin x}{1+x^2}\quad\text{so that}\quad \frac{dy}{dx}=y\left(\cos x\log\left(1+x^2\right)+\frac{2x\sin x}{1+x^2}\right).$$Substituting back for $y$ gives the result. 

Tell me if you don't understand any of the applications of the chain rule.
The rule that you quote is only valid if a is a constant.