How do you differentiate #f(x)= x * (4-x^2)^(1/2)# using the product rule?
1 Answer
Mar 5, 2016
Explanation:
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)#