Differentiate the functions with respect to x.

Let $f(x)=\sin \left(x^{2}+5\right), u(x)=x^{2}+5$, and $v(t)=\sin t$

Then, $($ vou $)=v(u(x))=v\left(x^{2}+5\right)=\tan \left(x^{2}+5\right)=f(x)$

Thus, f is a composite of two functions.

Put $t=u(x)=x^{2}+5$

Then, we obtain

$\frac{d v}{d t}=\frac{d}{d t}(\sin t)=\cos t=\cos \left(x^{2}+5\right)$

$\frac{d t}{d x}=\frac{d}{d x}\left(x^{2}+5\right)=\frac{d}{d x}\left(x^{2}\right)+\frac{d}{d x}(5)=2 x+0=2 x$

Therefore, by chain rule, $\frac{d f}{d x}=\frac{d v}{d t} \cdot \frac{d t}{d x}=\cos \left(x^{2}+5\right) \times 2 x=2 x \cos \left(x^{2}+5\right)$

Alternate method

$\frac{d}{d x}\left[\sin \left(x^{2}+5\right)\right]=\cos \left(x^{2}+5\right) \cdot \frac{d}{d x}\left(x^{2}+5\right)$

$=\cos \left(x^{2}+5\right) \cdot\left[\frac{d}{d x}\left(x^{2}\right)+\frac{d}{d x}(5)\right]$

$=\cos \left(x^{2}+5\right) \cdot[2 x+0]$

$=2 x \cos \left(x^{2}+5\right)$