# How do you differentiate f(x) = (2 x^2 +5 x +10)^cos(x)?

Jun 4, 2016

$f ' \left(x\right) = \frac{{\left(2 {x}^{2} + 5 x + 10\right)}^{\cos} \left(x\right) \left(\left(4 x + 5\right) \cos \left(x\right) - \left(2 {x}^{2} + 5 x + 10\right) \sin \left(x\right) \ln \left(2 {x}^{2} + 5 x + 10\right)\right)}{2 {x}^{2} + 5 x + 10}$

#### Explanation:

Let $y = {\left(2 {x}^{2} + 5 x + 10\right)}^{\cos} \left(x\right)$.

Then

$\ln \left(y\right) = \ln \left({\left(2 {x}^{2} + 5 x + 10\right)}^{\cos} \left(x\right)\right)$

$\ln \left(y\right) = \cos \left(x\right) \ln \left(2 {x}^{2} + 5 x + 10\right)$

Differentiating both sides, using the chain rule on the left and the product rule on the right:

$\frac{1}{y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = - \sin \left(x\right) \ln \left(2 {x}^{2} + 5 x + 10\right) + \cos \left(x\right) \left(\frac{4 x + 5}{2 {x}^{2} + 5 x + 10}\right)$

$\frac{1}{y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = \frac{\left(4 x + 5\right) \cos \left(x\right) - \left(2 {x}^{2} + 5 x + 10\right) \sin \left(x\right) \ln \left(2 {x}^{2} + 5 x + 10\right)}{2 {x}^{2} + 5 x + 10}$

Multiply both sides by $y$, or ${\left(2 {x}^{2} + 5 x + 10\right)}^{\cos} \left(x\right)$,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{\left(2 {x}^{2} + 5 x + 10\right)}^{\cos} \left(x\right) \left(\left(4 x + 5\right) \cos \left(x\right) - \left(2 {x}^{2} + 5 x + 10\right) \sin \left(x\right) \ln \left(2 {x}^{2} + 5 x + 10\right)\right)}{2 {x}^{2} + 5 x + 10}$