Derivative of $ f(x)=(\frac{x^2}{3x-1})^4(2x-3)^2$

Question

I determined $\Large f'(x)=((\frac{x^2}{3x-1})^4(2x-3)^2)'$ but the solution I'm given is different than what I got and I dont understand why. Here is what I worked out:

$\Large f'(x)=((\frac{x^2}{3x-1})^4(2x-3)^2)'$

$\Large = 4(\frac{x^2}{3x-1})^3(\frac{x^2}{3x-1})'(2x-3)^2+((2x-3)^2)'(\frac{x^2}{3x-1})^4$

$\Large = 4(\frac{x^2}{3x-1})^3(\frac{(x^2)'(3x-1)-(x^2)(3x-1)'}{(3x-1)^2})(2x-3)^2+(2(2x-3)(2x-3)')(\frac{x^2}{3x-1})^4$

$\Large = 4(\frac{x^2}{3x-1})^3(\frac{(2x)(3x-1)-(x^2)(3)}{(3x-1)^2})(2x-3)^2+(8x-12)(\frac{x^2}{3x-1})^4$

$\Large = 4(\frac{x^2}{3x-1})^3(\frac{(2x)(3x-1)-(x^2)(3)}{(3x-1)^2})(2x-3)^2+(8x-12)(\frac{x^2}{3x-1})^4$

I kept simplifying but never got to the given solution wich is:

$\Large 4(2x-3)\frac{9x^3 -14x^2+6x}{(3x-1)^2}(\frac{x^2}{3x-1})^3$

What am I doing wrong?

Answer 1

$((2x-3)^2)' = 2(2x-3)(2x-3)' = 4(2x-3) \ne 4(2x+3)$

Answer 2

Hint: I would simplify your term into $$f(x)=\frac{4 x^7 (2 x-3) \left(9 x^2-14 x+6\right)}{(3 x-1)^5}$$.

Answer 3

In such questions it is a huge relief if you take logarithms on both sides of the equations and then solve it. $$logf(x) = log(\frac{x^2}{3x-1})^{4}(2x-3)^2$$ $$logf(x) = 8log(x) -4log(3x-1)+2log(2x-3)$$ Now differentiating both sides with respect to x you will get : $$\frac{f'(x)}{f(x)} = \frac{8}{x}- \frac{4}{3x-1}*3 +\frac{2}{2x-3}*2$$ Take LCM and you will get : $$\frac{f'(x)}{f(x)} = \frac{8(3x-1)(2x-3)-12x(2x-3)+4x(3x-1)}{x(2x-3)(3x-1)}$$

Solve to get :

$$\frac{f'(x)}{f(x)} = \frac{4(9x^2-14x+6)}{x(2x-3)(3x-1)}$$ $$f'(x) = \frac{4(9x^2-14x+6)}{x(2x-3)(3x-1)}*(\frac{x^2}{3x-1})^4(2x-3)^2$$ The answer is : $$f'(x) = \frac{4(2x-3)(9x^3-14x^2+6x)}{(3x-1)^2}*(\frac{x^2}{3x-1})^3$$ Assuming that taking logarithms are in accordace with the rule that the function inside the log is positive and the base of the logarithm can be any real number. My favourite is e.