find the derivative of the function $f$ at $x=2$

Question

find the derivative of the function $f$ at $x=2$

$$f(x)=\dfrac{3(\sqrt[3]{5x-2})^2}{(2x-3)^5(x-1)^4}$$

My Try :

$$\lim_{x \to 2}\dfrac{\dfrac{3(\sqrt[3]{5x-2})^2}{(2x-3)^5(x-1)^4}-\dfrac{12(2x-3)^5(x-1)^4}{(2x-3)^5(x-1)^4}}{x-2}=\lim_{x\to2}\dfrac{3(\sqrt[3]{5x-2})^2-(24x-36)(2x^2-5x+3)^4}{(2x-3)^5(x-1)^4(x-2)}$$

Now what do i do ?

Answer 1

You should apply the rules of differentiation of products, quotients, compound functions... that you have learnt.

For efficiency, I'll use the "logarithmic derivative" trick:

$$(\log f(x))'=\dfrac{f'(x)}{f(x)}=\left(\log3+\frac23\log(5x-2)-5\log(2x-3)-4\log(x-1)\right)'$$

and

$$f'(x)=f(x)\left(\frac23\frac5{5x-2}-5\frac2{2x-3}-4\frac1{x-1}\right).$$

Now, evaluate at $x=2$.

Answer 2

Following your method, you should calculate

$$\lim_{x\to2} \frac{f(x)-f(2)}{x-2}$$

or as an alternative

$$\lim_{h\to0} \frac{f(2+h)-f(2)}{h}$$

Maybe in this case is more convenient differentiate and find directly $f'(x)$.