# How do you differentiate f(x)=cos(e^(3x^2)+7)  using the chain rule?

Jul 10, 2016

${f}^{'} = - 6 x {e}^{3 {x}^{2}} \sin \left({e}^{3 {x}^{2}} + 7\right)$

#### Explanation:

I prefer to use the form $\frac{\mathrm{dy}}{\mathrm{dx}}$

Let $u = {e}^{3 {x}^{2}} + 7 \implies \text{ } \frac{\mathrm{du}}{\mathrm{dx}} = 6 x {e}^{3 {x}^{2}}$

Write $f \left(x\right) \text{ as " y=cos(u)=>" } \frac{\mathrm{dy}}{\mathrm{du}} = - \sin \left(u\right)$

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But $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\cancel{\mathrm{du}}} \times \frac{\cancel{\mathrm{du}}}{\mathrm{dx}}$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \left(6 x {e}^{3 {x}^{2}}\right) \left(- \sin \left({e}^{3 {x}^{2}} + 7\right)\right)$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = - 6 x {e}^{3 {x}^{2}} \sin \left({e}^{3 {x}^{2}} + 7\right)$

Or if you prefer ${f}^{'} = - 6 x {e}^{3 {x}^{2}} \sin \left({e}^{3 {x}^{2}} + 7\right)$

As confirmation: