# How do you find dy/dt given y= 4 sin ( sqrt (1+ sqrt (x) ))?

Dec 24, 2017

The chain rule says:

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

We can compute $\frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 4 \cos \left(\sqrt{1 + \sqrt{x}}\right) \frac{1}{2 \sqrt{1 + \sqrt{x}}} \frac{1}{2 \sqrt{x}}$

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

But, because we are not given a function of x in terms of t, we cannot compute $\frac{\mathrm{dx}}{\mathrm{dt}}$. Therefore, we must leave the answer in this from:

$\frac{\mathrm{dy}}{\mathrm{dt}} = \cos \frac{\sqrt{1 + \sqrt{x}}}{\sqrt{1 + \sqrt{x}} \sqrt{x}} \frac{\mathrm{dx}}{\mathrm{dt}}$