How do you differentiate  y =x sqrt((4-x^2)  using the chain rule?

Jan 21, 2016

$h ' \left(x\right) = \sqrt{4 - {x}^{2}} - \frac{1}{\sqrt{4 - {x}^{2}}} {x}^{2}$
$h ' \left(x\right) = - \frac{2 {x}^{2} - 4}{\sqrt{4 - {x}^{2}}}$

Explanation:

This actually a product of 2 functions:
Let
Use the product rule and write:

1) $h \left(x\right) = f \left(x\right) g \left(x\right) \mathmr{and} h ' \left(x\right) = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$

We use the chain rule on g(x):
Let $g \left(x\right) = \left(r \circ p\right) \left(x\right)$
then
2) $g ' \left(x\right) = r ' \left(p \left(x\right)\right) \cdot p ' \left(x\right)$
Now: r(x) =sqrt(x); p(x) = 4-x^2
r'(x) = 1/(2sqrt(x)); p'(x) = -2x
$g ' \left(x\right) = \frac{1}{\cancel{2} \sqrt{4 - {x}^{2}}} \cdot \cancel{\left(- 2\right) x}$

$g ' \left(x\right) = - \frac{1}{\sqrt{4 - {x}^{2}}} \cdot \left(x\right)$

So what we have is:
f(x) = x; f'(x)= 1
g(x) = sqrt(4-x^2); g'(x) =-1/(sqrt(4-x^2)) x

From 1)
$h ' \left(x\right) = \sqrt{4 - {x}^{2}} - \frac{1}{\sqrt{4 - {x}^{2}}} {x}^{2}$

Jan 21, 2016

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

Explanation:

$y = f \left(x\right) \cdot g \left(x\right)$

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

we have to use Product Rule and Chain Rule:

$y ' = f ' \left(x\right) \cdot g \left(x\right) + f \left(x\right) g ' \left(x\right)$

$g ' \left(x\right) = \frac{d}{\mathrm{dx}} h \left(i \left(x\right)\right) = h ' \left(i \left(x\right)\right) \cdot i ' \left(x\right)$

therefore:

$y ' = 1 \cdot \sqrt{\left(4 - {x}^{2}\right)} + x \cdot \left(\frac{1}{2 \sqrt{\left(4 - {x}^{2}\right)}} \cdot \left(0 - 2 x\right)\right) =$

$= \sqrt{\left(4 - {x}^{2}\right)} + x \cdot \left(\frac{1}{\cancel{2} \sqrt{\left(4 - {x}^{2}\right)}} \left(- \cancel{2} x\right)\right) =$
$= \sqrt{\left(4 - {x}^{2}\right)} - {x}^{2} / \sqrt{\left(4 - {x}^{2}\right)}$
$= \frac{4 - {x}^{2} - {x}^{2}}{\sqrt{4 - {x}^{2}}} = 2 \cdot \frac{2 - {x}^{2}}{\sqrt{4 - {x}^{2}}}$