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

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.

• I, myself, am familiar with the product rule, but even that is taught later in the book, so I would assume that I have to solve this without it.
– Liam
Aug 4, 2019 at 9:41
• Notice that it would be much easier using logarithmic differentiation (but this would be off-topic). Aug 4, 2019 at 13:26

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.

• I didn't know what to do after having to find the derivative of $x\div (1-\sqrt{x})$, so nice substitution! +1 :) Aug 4, 2019 at 11:26

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.

• In your 2nd line of work, would the last x not be to the power of negative half? Also, the product rule is being used in combination with the chain rule here. This question in the book comes before the product rule is taught, thus, I do not believe I can use it to solve it like this.
– Liam
Aug 4, 2019 at 9:51
• @Liam You're right about that mistake. I've fixed the typo. As for the product rule, I don't offhand see any particular way to avoid using it. However, I will give it more thought and see if I can think of any way. Aug 4, 2019 at 9:54
• @Liam I just realized I made a mistake with summing the square roots of $x$, so my answer now does agree with yours, although I expressed it somewhat differently. As for your idea about worked vs. non-worked solutions, this is definitely possible. Aug 4, 2019 at 10:02
• See the inversion trick in my answer below. Aug 4, 2019 at 10:18
• @TheSimpliFire It's a neat trick to avoid needing to use the product rule in this case. Aug 4, 2019 at 10:25

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 $$f'(x)=-\frac{3 \left(\sqrt{x}-2\right) x^2}{2 \left(\sqrt{x}-1\right)^4}$$

• Yes, it must be negative, corrected! Thank you! Aug 4, 2019 at 10:03

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}