How do you find the first derivative of y=(sinx/(1+cosx))^2?

May 22, 2015

Let's use the chain rule, by naming $u = \frac{\sin x}{1 + \cos x}$.

The chain rule states that

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

Thus,

$\frac{\mathrm{dy}}{\mathrm{du}} = 2 u$

$\frac{\mathrm{du}}{\mathrm{dx}}$ - here, we have to apply quocient rule, which states that

Be $y = \frac{f \left(x\right)}{g \left(x\right)}$, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{f \left(x\right)} ^ 2$

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{\left(\left(\cos x\right) \left(1 + \cos x\right)\right) - \left(\sin x\right) \left(- \sin x\right)}{1 + \cos x} ^ 2$

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{\left({\cos}^{2} x + \cos x\right) + {\sin}^{2} x}{1 + \cos x} ^ 2$

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{\textcolor{g r e e n}{{\cos}^{2} x + {\sin}^{2} x} + \cos x}{1 + \cos x} ^ 2$

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{\cancel{1 + \cos x}}{1 + \cos x} ^ \cancel{2}$

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{1 + \cos x}$

Now, combining $\frac{\mathrm{du}}{\mathrm{dx}}$ and $\frac{\mathrm{dy}}{\mathrm{du}}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(2 u\right) \left(\frac{1}{1 + \cos x}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \left(\sin \frac{x}{1 + \cos x}\right) \left(\frac{1}{1 + \cos x}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \textcolor{g r e e n}{\frac{2 \sin x}{1 + \cos x} ^ 2}$