Taking the derivative of $y = (\frac{x}{1-\sqrt{x}})^3$ using the chain rule

Question

While working through differential calculus questions for the chain rule, I stumbled upon:

$$y = \left(\frac{x}{1-\sqrt{x}}\right)^3 $$

I initially attempted to apply the chain rule, but to apply it, I would need to differentiate the contents in the brackets, which, from what I know, I can only differentiate using the quotient rule. However, in my book, the quotient rule is taught later and thus I would assume that I can only use the mathematical tools taught thus far, i.e. the chain rule and the 'differentiating short-cut' (that's what my teacher calls it), i.e. if $f(x) = ax^n$, $f'(x) = anx^{n-1}$. I cannot figure out a way to solve this question by only using only the chain rule and differentiating short-cut; am I missing something or is the question simply in the wrong place in my book?

I would also like to point out that when I looked at the worked solutions for this maths book, they had all answers for this exercise except for that question. The solutions displayed without working is:

$$\frac{1-2\sqrt{x}}{4\sqrt{x-x\sqrt{x}}}$$

Even if I were to use the quotient rule and chain rule, I get a different answer (I even repeated my working twice in case I made a mistake, but I got the same answer both times):

$$\frac{3x^2-\frac{3}{2}x^{\frac{5}{2}}}{(1-\sqrt{x})^4}$$

EDIT: I believe the product rule can also not be used to solve this question as it is, just like the quotient rule, taught later in the book.

Bibliography:

Mathematics Higher Level, IB, by Josip Harcet et. al.

Answer 1

Solving using only the chain rule

From the chain rule, $$y = \left(\frac{x}{1-\sqrt{x}}\right)^3\implies\frac{dy}{dx}=3\left(\frac x{1-\sqrt x}\right)^2\cdot\color{red}{\frac d{dx}\left(\frac{x}{1-\sqrt{x}}\right)}\tag1.$$ Now \begin{align}\color{red}{\frac d{dx}\left(\frac{x}{1-\sqrt{x}}\right)}&=\frac d{dx}\left(\left(\frac{1-\sqrt x}{x}\right)^{-1}\right)=-\left(\frac{1-\sqrt x}{x}\right)^{-2}\cdot\color{blue}{\frac d{dx}\left(\frac{1-\sqrt x}{x}\right)}\tag2\end{align} and $$\color{blue}{\frac d{dx}\left(\frac{1-\sqrt x}{x}\right)}=\frac d{dx}\left(x^{-1}-x^{-1/2}\right)=-\frac1{x^2}+\frac1{2x\sqrt x}\tag3$$ so \begin{align}\frac{dy}{dx}&=3\left(\frac x{1-\sqrt x}\right)^2\cdot\left(-\left(\frac{1-\sqrt x}{x}\right)^{-2}\right)\cdot\left(-\frac1{x^2}+\frac1{2x\sqrt x}\right)\\&=-3\left(\frac x{1-\sqrt x}\right)^2\cdot\left(\frac x{1-\sqrt x}\right)^2\cdot\left(-\frac1{x^2}+\frac1{2x\sqrt x}\right)\\&=\frac{3x^4}{(1-\sqrt x)^4}\left(\frac1{x^2}-\frac1{2x\sqrt x}\right)\\\vphantom{2cm}\\\implies\frac{dy}{dx}&=\frac{3x^2-\frac32x^{5/2}}{(1-\sqrt x)^4}\tag4\end{align} which is what you have. As for the book's answer, it's wrong, as can be seen here.

Answer 2

A quotient can be expressed as a product of the function to the power of $-1$, e.g., $\frac{f(x)}{g(x)} = f(x)g^{-1}(x)$. You can then apply your "differentiating short-cut" using $n = -1$. In particular, this gives $h(x) = g^{-1}(x) \implies h'(x) = -g^{-2}(x)$, which is the same as the quotient rule.

As for the specific question,

$$\begin{equation}\begin{aligned} f(x) & = x^3\left(1-\sqrt{x}\right)^{-3} \\ f'(x) & = 3x^2\left(1-\sqrt{x}\right)^{-3} + x^3(-3)\left(1-\sqrt{x}\right)^{-4}\left(-\frac{1}{2}x^{-1/2}\right) \\ & = 3x^2\left(1-\sqrt{x}\right)\left(1-\sqrt{x}\right)^{-4} + \frac{3}{2}x^{5/2}\left(1-\sqrt{x}\right)^{-4} \\ & = \frac{3x^2\left(1 - \sqrt{x} + \frac{1}{2}\sqrt{x}\right)}{\left(1-\sqrt{x}\right)^{4}} \\ & = \frac{3x^2\left(1 - \frac{1}{2}\sqrt{x}\right)}{\left(1-\sqrt{x}\right)^{4}} \\ & = \frac{3x^2\left(2 - \sqrt{x}\right)}{2\left(1-\sqrt{x}\right)^{4}} \\ \end{aligned}\end{equation}\tag{1}\label{eq1}$$

This basically agrees with what you got. As for the maths book solution, I don't see how they got that.

Answer 3

We have $$f(x)=\frac{x^3}{(1-\sqrt{x})^3}=x^3\cdot (1-\sqrt{x})^{-3}$$ So we get $$f'(x)=3x^2\cdot (1-\sqrt{x})^{-3}+x^3\cdot (-3)(1-\sqrt{x})^{-4}\cdot (-1)\frac{1}{2}x^{-1/2}$$ It can be simplified to $$f'(x)=-\frac{3 \left(\sqrt{x}-2\right) x^2}{2 \left(\sqrt{x}-1\right)^4}$$

Answer 4

Alternatively, using "the chain rule and differentiating short-cut": $$\begin{align}y &= \left(\frac{x}{1-\sqrt{x}}\right)^3= \left(\frac{1-\sqrt{x}}{x}\right)^{-3}=(x^{-1}-x^{-1/2})^{-3};\\ y'&=-3(x^{-1}-x^{-1/2})^{-4}\cdot (-x^{-2}+\frac12x^{-3/2})=\\ &=-3\left(\frac1x-\frac1{\sqrt{x}}\right)^{-4}\cdot (-\frac1{x^2}+\frac1{2x^{3/2}})=\\ &=3\left(\frac{1-\sqrt{x}}{x}\right)^{-4}\cdot \frac{2-\sqrt{x}}{2x^2}=\\ &=3\left(\frac{x}{1-\sqrt{x}}\right)^4\cdot \frac{2-\sqrt{x}}{2x^2}=\\ &=\frac{3x^2(2-\sqrt{x})}{(1-\sqrt{x})^4}\end{align}$$