# What is the derivative of sin(x^2+5) cos(x^2+9x+2)?

Sep 9, 2015

$x \cdot \cos \left({x}^{2} + 5\right) \cdot \cos \left({x}^{2} + 9 x + 2\right) - \left(2 x + 9\right) \cdot \sin \left({x}^{2} + 5\right) \cdot \sin \left({x}^{2} + 9 x + 2\right)$

#### Explanation:

You can differentiate this function by using the product rule

color(blue)(d/dx(f(x) * g(x)) = [d/dx(f(x))] * g(x) + f(x) * d/dx(g(x))

the fact that

$\frac{d}{\mathrm{dx}} \left(\sin x\right) = \cos x \text{ }$ and $\text{ } \frac{d}{\mathrm{dx}} \left(\cos x\right) = - \sin x$

and the chain rule for $\sin u$, with $u = {x}^{2} + 5$ and $\cos t$, with $t = {x}^{2} + 9 x + 2$.

So, put all this together to get

$y = \sin u \cdot \sin t$

$\frac{d}{\mathrm{dx}} \left(y\right) = \left[\frac{d}{\mathrm{dx}} \left(\sin u\right)\right] \cdot \cos t + \sin u \cdot \frac{d}{\mathrm{dx}} \left(\cos t\right)$

${y}^{'} = \left[\frac{d}{\mathrm{du}} \left(\sin u\right) \cdot \frac{d}{\mathrm{dx}} \left(u\right)\right] \cdot \cos t + \sin u \cdot \left[\frac{d}{\mathrm{dt}} \left(\cos t\right) \cdot \frac{d}{\mathrm{dx}} \left(t\right)\right]$

${y}^{'} = \left[\cos u \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} + 5\right)\right] \cdot \cos t + \sin u \cdot \left[\left(- \sin t\right) \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} + 9 x + 2\right)\right]$

${y}^{'} = \left[\cos \left({x}^{2} + 5\right) \cdot 2 x\right] \cdot \cos \left({x}^{2} + 9 x + 2\right) + \sin \left({x}^{2} + 5\right) \cdot \left[- \sin \left({x}^{2} + 9 x + 2\right) \cdot \left(2 x + 9\right)\right]$

${y}^{'} = \textcolor{g r e e n}{2 x \cdot \cos \left({x}^{2} + 5\right) \cdot \cos \left({x}^{2} + 9 x + 2\right) - \left(2 x + 9\right) \cdot \sin \left({x}^{2} + 5\right) \cdot \sin \left({x}^{2} + 9 x + 2\right)}$