# How do you use the chain rule to differentiate y=sin^3x+cos^3x?

Oct 7, 2016

#### Explanation:

$y = {\sin}^{3} x + {\cos}^{3} x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left({\sin}^{3} x\right) + \frac{d}{\mathrm{dx}} \left({\cos}^{3} x\right)$

Now we'll be using the chain rule to get $\frac{\mathrm{dy}}{\mathrm{dx}}$...

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

So, here it goes...

$\frac{d}{\mathrm{du}} \left({u}^{3}\right) = 3 {u}^{2} = 3 {\sin}^{2} x$

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

$\frac{d}{\mathrm{du}} \left({u}^{3}\right) \cdot \frac{d}{\mathrm{dx}} \left(\sin x\right) = 3 {\sin}^{2} x \cos x$

Which means that:

$\frac{d}{\mathrm{dx}} \left({\sin}^{3} x\right) = 3 {\sin}^{2} x \cos x$

Now...

$\frac{d}{\mathrm{dp}} \left({p}^{3}\right) = 3 {p}^{2} = 3 {\cos}^{2} x$

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

$\frac{d}{\mathrm{dp}} \left({p}^{3}\right) \cdot \frac{d}{\mathrm{dx}} \left(\cos x\right) = - 3 {\cos}^{2} x \sin x$

Which means that...

$\frac{d}{\mathrm{dx}} \left({\cos}^{3} x\right) = - 3 {\cos}^{2} x \sin x$

Since...

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left({\sin}^{3} x\right) + \frac{d}{\mathrm{dx}} \left({\cos}^{3} x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {\sin}^{2} x \cos x - 3 {\cos}^{2} x \sin x$

$= 3 \sin x \cos x \left(\sin x - \cos x\right)$

You can then again transform this result if you wish to.