How do you differentiate  y =-(e^(x-sin^2x) using the chain rule?

Oct 9, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \left(1 - \sin 2 x\right) {e}^{x - {\sin}^{2} x}$

Explanation:

In order to differentiate a function of a function, say $y , = f \left(g \left(x\right)\right)$, where we have to find $\frac{\mathrm{dy}}{\mathrm{dx}}$, we need to do (a) substitute $u = g \left(x\right)$, which gives us $y = f \left(u\right)$. Then we need to use a formula called Chain Rule, which states that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$. In fact if we have something like $y = f \left(g \left(h \left(x\right)\right)\right)$, we can have $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{df}} \times \frac{\mathrm{df}}{\mathrm{dg}} \times \frac{\mathrm{dg}}{\mathrm{dh}}$

Hence for $y = - {e}^{x - {\sin}^{2} x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {e}^{x - {\sin}^{2} x} \times \frac{d}{\mathrm{dx}} \left(x - {\sin}^{2} x\right)$

= $- {e}^{x - {\sin}^{2} x} \times \left(1 - 2 \sin x \times \frac{d}{\mathrm{dx}} \sin x\right)$

= $- {e}^{x - {\sin}^{2} x} \times \left(1 - 2 \sin x \times \cos x\right)$

= $- \left(1 - \sin 2 x\right) {e}^{x - {\sin}^{2} x}$