How do you find the derivative of x*(sqrt(4-x^2))?

Mar 23, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 {x}^{2}}{2 \sqrt{4 - {x}^{2}}} + \sqrt{4 - {x}^{2}}$

Explanation:

Before offering a solution to the problem, let me tell you an easy method of finding the derivative of $\sqrt{x}$

You put the derivative of $x$ in the numerator and multiply the given function with 2 and put it in the denominator.

The derivative of $\sqrt{x} = \frac{1}{2 \sqrt{x}}$

Use this simple technique in the given problem

==================

Now the problem

Given -

$y = x . \left(\sqrt{4 - {x}^{2}}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = x . \left[\frac{- 2 x}{2 \sqrt{4 - {x}^{2}}}\right] + \sqrt{4 - {x}^{2}} . \left(1\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 {x}^{2}}{2 \sqrt{4 - {x}^{2}}} + \sqrt{4 - {x}^{2}}$

Mar 23, 2017

$f ' \left(x\right) = \frac{4 - 2 {x}^{2}}{\sqrt{4 - {x}^{2}}}$

Explanation:

$f \left(x\right) = x \cdot \sqrt{4 - {x}^{2}}$

$f ' \left(x\right) = x \cdot \frac{d}{\mathrm{dx}} {\left(4 - {x}^{2}\right)}^{\frac{1}{2}} + {\left(4 - {x}^{2}\right)}^{\frac{1}{2}} \cdot 1$ [Product rule]

$= x \cdot \frac{1}{2} \cdot {\left(4 - {x}^{2}\right)}^{- \frac{1}{2}} \cdot \left(- 2 x\right) + {\left(4 - {x}^{2}\right)}^{\frac{1}{2}}$ [Chain rule]

$= \frac{x \cdot - 2 x}{2 \sqrt{4 - {x}^{2}}} + \sqrt{4 - {x}^{2}}$

$= \sqrt{4 - {x}^{2}} - {x}^{2} / \sqrt{4 - {x}^{2}}$

$= \frac{4 - {x}^{2} - {x}^{2}}{\sqrt{4 - {x}^{2}}}$

$= \frac{4 - 2 {x}^{2}}{\sqrt{4 - {x}^{2}}}$