Take the derivative of the function $f(x)=(\frac{8x^2-5}{2-x})^3$

Question

Section 2.5 #10:

Problem: Take the derivative of the function $f(x)=(\frac{8x^2-5}{2-x})^3$

Solution: Taking this derivative will involve the chain rule AND the quotient rule. First we use the chain rule:

$f'(x)=\frac{d}{dx}(\frac{8x^2-5}{2-x})^3$

$f'(x)=3(\frac{8x^2-5}{2-x})^2(\frac{d}{dx}(\frac{8x^2-5}{2-x}))$

Side quest: evaluate $\frac{d}{dx}(\frac{8x^2-5}{2-x})$. We are going to use the quotient rule here.

$\frac{d}{dx}(\frac{8x^2-5}{2-x})$

$= \frac{(2-x)\frac{d}{dx}(8x^2-5)-(8x^2-5)\frac{d}{dx}(2-x)}{(2-x)^2}$

$= \frac{(2-x)(16x-0)-(8x^2-5)(0-1)}{(2-x)^2}$

$= \frac{(2-x)(16x)-(8x^2-5)(-1)}{(2-x)^2}$

When you do the quotient rule, be very careful to distribute the negative sign you get when subtracting correctly, this is where many people screw up!!! Note here, in this example, we also have the $-1$ to distribute as well, making keeping track of whether things are positive or negative even more complicated.

$= \frac{32x-16x^2+8x^2-5}{(2-x)^2}$

$= \frac{32x-8x^2-5}{(2-x)^2}$

Cool... Now we plug the value we just calculated of

$\frac{d}{dx}(\frac{8x^2-5}{2-x})$

back into $f'(x)=3(\frac{8x^2-5}{2-x})^2(\frac{d}{dx}(\frac{8x^2-5}{2-x}))$

to get

$f'(x)=3(\frac{8x^2-5}{2-x})^2(\frac{32x-8x^2-5}{(2-x)^2})$

This is an acceptable answer. We could however write it a little nicer using the properties of exponents:

$f'(x)=3(\frac{(8x^2-5)^2}{(2-x)^2})(\frac{32x-8x^2-5}{(2-x)^2})$

$f'(x)=\frac{3(32x-8x^2-5)(8x^2-5)^2}{(2-x)^4}$

Answer 1

Your question has not been clearly stated. In any case, you can avoid the complication of using quotient rule by performing division of the rational expression (it's easy to use synthetic division).

The expression is $(\frac{8x^2-5}{2-x})^3=(\frac{-8x^2+5}{x-2})^3= (-8x-16-\frac{27}{x-2})^3$

and its derivative is $3(-8x-16-\frac{27}{x-2})^2(-8+\frac{27}{(x-2)^2})$

which can be equivalently written as $3(\frac{8x^2-5}{2-x})^2(\frac{27}{(x-2)^2}-8)$

and you can simplify further.

Answer 2

One useful trick when facing products, quotients, powers, ... is logarithmic differentiation. $$f(x)=\left(\frac{8x^2-5}{2-x}\right)^3 \implies \log(f(x))=3 \log(8x^2-5)-3\log(2-x)$$ $$\frac{f'(x)}{f(x)}=3 \frac{16x}{8x^2-5}-3 \frac {-1}{2-x}=3\frac{8 x^2-32 x+5 }{(8x^2-5)(2-x) }$$ Now $$f'(x)=f(x) \times \frac{f'(x)}{f(x)}=\frac{(8x^2-5)^3}{(2-x)^3}\times3\times\frac{8 x^2-32 x+5 }{(8x^2-5)(2-x) }$$

Simplify.