How do you differentiate #g(x) = x^2sqrt(2x+1)# using the product rule?
1 Answer
Feb 19, 2016
Explanation:
The Product Rule:
#frac{"d"}{"d"x}(uv) = v*frac{"d"u}{"d"x} + u*frac{"d"v}{"d"x}#
Therefore,
#g'(x) = frac{"d"}{"d"x}(x^2sqrt(2x+1))#
#= x^2*frac{"d"}{"d"x}(sqrt(2x+1)) + sqrt(2x+1)*frac{"d"}{"d"x}(x^2)#
#= x^2*2/(2sqrt(2x+1)) + sqrt(2x+1)*(2x)#
#= x^2/sqrt(2x+1) + frac{2x*(2x+1)}{sqrt(2x+1)}#
#= frac{5x^2+2x}{sqrt(2x+1)}#
