Chain rule of fractions

Question

How can I use chain rule to find the derivative of $y=\frac{(3x-3)^2}{x}$?
Should I multiply the derivative of whole $y$ with the derivative of $(3x-3)^2$ and then after that all with derivative of $(3x-3)$?

Answer 1

Firstly, you have a rational function. So, you have to consider the product rule or quotient rule. Let us use the quotient rule (as C. Falcon has pointed out): $$\frac{d}{dx}\biggl(\frac{(3x-3)^2}{x}\biggl) = \frac{6x*(3x-3) - (3x-3)^2}{x^2}$$

The chain rule was when we were differentiating $(3x-2)^2$.

Answer 2

Better apply the quotient rule. To derivate the numerator, you also need the chain rule. I do not see a way only using the chain rule.

Answer 3

Let $\iota\colon\mathbb{R}^*\rightarrow\mathbb{R}$ be the inverse function defined by: $$\iota(x):=\frac{1}{x}.$$ I assume you know that one has: $$\iota'(x)=-\frac{1}{x^2}.$$ Let $u,v\colon\mathbb{R}\rightarrow\mathbb{R}$ differentiable maps with $v$ non-vanishing and let $f:=u/v=u\times\iota\circ v$.

Using product rule first, one has: $$f'=u'(\iota\circ v)+u(\iota\circ v)'.$$ Besides, using chain rule, one has: $$(\iota\circ v)'=v'\iota'\circ v=-v'\frac{1}{v^2}.$$ Therefore, putting things together, one gets: $$f'=\frac{u'v-v'u}{v^2}.$$

Answer 4

As an alternative, you can also derive by the product rule:

$$y=\frac{(3x-3)^2}{x}=f(x)\cdot g(x)$$

with:

$$f(x)=(3x-3)^2$$

$$g(x)=\frac{1}{x}$$ then: $$y'=f'(x)g(x)+f(x)g'(x)$$

Answer 5

With stuff like this you can also expand it to $f(x)=9x-18+\frac 9x$ and derivate $f'(x)=9-\frac 9{x^2}$, this is more efficient.

However if you have calculus withdrawal symptoms already you can either use:

• The product rule : $(uv)'=u'v+v'u$

$f'(x)=\underbrace{6(3x-3)}_{u'}\times \underbrace{\dfrac 1x}_{v} + \underbrace{(3x-3)^2}_{u}\times \underbrace{\dfrac{-1}{x^2}}_{v'}$

• Or the quotient rule: $(\frac uv)'=\dfrac{u'v-v'u}{v^2}$

$f'(x)=\dfrac{\underbrace{6(3x-3)}_{u'}\times \underbrace{x}_{v}-\underbrace{1}_{v'}\times\underbrace{(3x-3)^2}_{u}}{\underbrace{x^2}_{v^2}}$