How do you differentiate #f(x)= x * (4-x^2)^(1/2)# using the product rule?

Question

Paisley Johnson · Accepted Answer

#f'(x) = (4 - 2x^2)/sqrt(4 - x^2)#

The Product Rule:

#frac{"d"}{"d"x}(uv) = v frac{"d"u}{"d"x} + u frac{"d"v}{"d"x}#

In this question,

#u = x#

#v = (4 - x^2)^{1/2}#

#f'(x) = frac{"d"}{"d"x}(x(4 - x^2)^{1/2})#

#= (4 - x^2)^{1/2} frac{"d"}{"d"x}(x) + x frac{"d"}{"d"x}((4 - x^2)^{1/2})#

#= (4 - x^2)^{1/2} (1) + x (1/2) (4 - x^2)^{-1/2} frac{"d"}{"d"x}(4 - x^2)#

#= sqrt(4 - x^2) + x (1/2) (4 - x^2)^{-1/2} (- 2x)#

#= sqrt(4 - x^2) - x^2/sqrt(4 - x^2)#

#= ((sqrt(4 - x^2))^2 - x^2)/sqrt(4 - x^2)#

#= (4 - 2x^2)/sqrt(4 - x^2)#

Christopher Allers · Answer

undefined To differentiate $ f(x) = x \cdot (4 - x^2)^{\frac{1}{2}} $ using the product rule, let $ u = x $ and $ v = (4 - x^2)^{\frac{1}{2}} $. Then, $ u' = 1 $ and $ v' = -\frac{x}{\sqrt{4 - x^2}} $. Applying the product rule, we get:

$$ f'(x) = u'v + uv' $$ 
$$ f'(x) = (1) \cdot (4 - x^2)^{\frac{1}{2}} + (x) \cdot \left( -\frac{x}{\sqrt{4 - x^2}} \right) $$ 
$$ f'(x) = (4 - x^2)^{\frac{1}{2}} - \frac{x^2}{\sqrt{4 - x^2}} $$