How do you differentiate #f(x)=2 ln( x^2 - 3x +4) # using the chain rule?

1 Answer

Answer:

#f'(x)=\frac{2(2x-3)}{x^2-3x+4}#

Explanation:

Given function:

#f(x)=2\ln(x^2-3x+4)#

Differentiating above function w.r.t. #x# using chain rule as follows

#\frac{d}{dx}f(x)=\frac{d}{dx}2\ln(x^2-3x+4)#

#f'(x)=2\frac{d}{dx}\ln(x^2-3x+4)#

#=2(\frac{1}{x^2-3x+4})\frac{d}{dx}(x^2-3x+4)#

#=\frac{2}{x^2-3x+4}(2x-3)#

#=\frac{2(2x-3)}{x^2-3x+4}#