# how to solve a derivative with the quotient rule

I have a function that is

$f(x) = \frac{(10x^2)(x^2-3)}{(x^2 -1)^2}$

I need to find the derivative of this function, and I simplify the numerator so the equation looks like $f(x) = \frac{(10x^4-30x^2)}{(x^2 - 1)^2}$. I am using the derivative quotient rule to solve the entire problem and I use the derivative product rule in the numerator.

So I do $f'(x) = \frac{\frac{d}{dx}(10x^4-30x^2)*(x^2 -1)^2 - (10x^4 - 30x^2)\frac{d}{dx}(x^2 - 1)^2}{(x^2 - 1)^4}$

I apply the chain rule to the $\frac{d}{dx}(x^2 - 1)^2$ but I end up with $40x^3+10x^4-90x^2-40x^7+160x^5-120x^3$ in the numerator, which is wrong because the answer in the book the answer for the equation is $f'(x) = \frac{20x(x^2-3)}{(x^2 -1)^3}$. Can someone help me figure out what I am doing wrong?

• First off, I wouldn't multiply out the numerator until after you have killed off one of those factors of $(x^2-1)$. That should make your life a little easier. Mar 26, 2018 at 23:54

$\frac {d}{dx}( \frac{(10x^2)(x^2-3)}{(x^2 -1)^2})\\ \frac {(x^2 - 1)^2\left((20x)(x^2 -3) + (10x^2)(2x)\right) - \left(2(x^2 - 1)(2x)\right)(10x^2)(x^2 - 3)}{(x^2-1)^4}$

Every term in the numerator has a factor of $(x^2 - 1)$ that cancels with the denominator.

$\frac {(x^2 - 1)\left((20x)(x^2 -3) + (10x^2)(2x)\right) - (4x)(10x^2)(x^2 - 3)}{(x^2-1)^3}$

And every term has a $20x$ factor, which we can take to the front.

$\frac {20x \left( (x^2 - 1)(2x^2 -3) - (2x^2)(x^2 - 3)\right)}{(x^2-1)^3}$

Multiplying out the numerator and simplify. It looks like the $x^4$ terms cancel.

$\frac {20x( 2x^4 - 5x^2 + 3 - 2x^4 + 6x^2)}{(x^2-1)^3}\\ \frac {20x(x^2 + 3)}{(x^2-1)^3}\\$

We have

$$f(x)=\frac{(10x^2)(x^2-3)}{(x^2 -1)^2}=\frac{10x^4-30x^2}{(x^2 -1)^2}$$

then

$$f'(x)=\frac{(40x^3-60x)(x^2-1)^2-4x(x^2-1)(10x^4-30x^2)}{(x^2 -1)^4} =\frac{(40x^3-60x)(x^2-1)-4x(10x^4-30x^2)}{(x^2 -1)^3} =\frac{40x^5-100x^3+60x-40x^5+120x^3}{(x^2 -1)^3} =\frac{20x^3+60x}{(x^2 -1)^3}=\frac{20x(x^2+3)}{(x^2 -1)^3}$$

Since there’s a product in the numerator, I’d use a generalization of the product rule instead of the quotient rule: $$(fgh)' = f'gh+fg'h+fgh'.$$ Taking $$f(x)=10x^2 \\ g(x)=x^2-3 \\ h(x)=(x^2-1)^{-2}$$ we have $$f'(x)=20x \\ g'(x)=2x \\ h'(x) = -{4x\over(x^2-1)^3}.$$ Putting this all together, \begin{align} (fgh)' &= {(20x)(x^2-3)\over(x^2-1)^2} + {(10x^2)(2x)\over(x^2-1)^2} + {(10x^2)(x^2-3)(-4x)\over(x^2-1)^3} \\ &= {20x(x^2-3)(x^2-1)+20x^3(x^2-1)-40x^3(x^2-3) \over (x^2-1)^3} \\ &= {20x(x^2+3) \over (x^2-1)^3}.\end{align}

For this kind of problems with only powers, products and quotients, logarithmic differentiation makes life much easier. $$y = \frac{(10x^2)(x^2-3)}{(x^2 -1)^2}\implies \log(y)=\log(10)+2\log(x)+\log(x^2-3)-2\log(x^2-1)$$ Differentiate both sides $$\frac{y'}y=\frac 2x+\frac{2x}{x^2-3}-\frac{4x}{x^2-1}$$ Simplify as much as you can and, at the end, use $$y'=y \times \left(\frac{y'}y\right)$$ and continue simplifications.