How to get first derivative?

Question

My question is to find the first derivative using product and/or quotient rule

$$f(x) = {(x^2+1)(x^2-2) \over 3x+2}.$$

The solution to the problem is: $$f'(x) = {[(2x)(x^2-2)+(x^2+1)(2x)](3x+2)-(3)[(x^2+1)(x^2-2)] \over (3x+2)^2}.$$

I'm having problems getting the same solution would someone be able to help me by showing me the correct way of getting the first derivative of this question

Answer 1

HINT

Let apply the following

$$\left(\frac{f(x)}{g(x)}\right)'=\frac{f'(x)\cdot g(x)-g'(x)f(x)}{g^2(x)}$$

with $$f(x)=p(x)\cdot q(x) \implies f'(x)=p'(x)\cdot q(x)+p(x)\cdot q'(x)$$

Answer 2

For those who don't like differentiating fractions, the following uses just the product rule once:

$$ \begin{align} (3x+2)f(x) = x^4 -x^2-2 \;\;&\implies\;\; 3f(x)+(3x+2)f'(x)=4x^3-2x \\[5px]&\implies\;\;f'(x) = \dfrac{1}{3x+2}\left(4x^3-2x-3f(x)\right) \;= \;\ldots \end{align} $$

Answer 3

When you have expressions which just contains products, quotients and powers, logarithmic differentiation makes life easier.

For your case $$f = {(x^2+1)(x^2-2) \over 3x+2}\implies \log(f)=\log(x^2+1)+\log(x^2-2)-\log(3x+2)$$ Differentiate both sides $$\frac{f'}f=\frac{2x}{x^2+1}+\frac{2x}{x^2-2}-\frac 3{3x+2}=\frac{9 x^4+8 x^3-3 x^2-4 x+6 }{(x^2+1 )(x^2-2 ) (3x+2 ) }$$ Now $$f'=f \times \frac{f'}f={(x^2+1)(x^2-2) \over 3x+2}\times \frac{9 x^4+8 x^3-3 x^2-4 x+6 }{(x^2+1 )(x^2-2 ) (3x+2 ) }$$ Now, simplify.