Finding the derivative of $y=\frac{1}{\left(x+\sqrt{4+\left(x^2-1\right)^3}\right)^4}$

Question

Find an expression for $\frac{dy}{dx}$ given that

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

To be honest, I don't know how hard this question is supposed to be as it was found in an A-level textbook or if there is a simpler method to use than the one I did. I am fairly sure however that this question would never appear on a standard A-level maths paper. I used the chain rule but there were stages in my working where I'm not quite sure what I'm doing is correct.

The textbook says that the answer is

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

But I am getting

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

and online derivative calculators say

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

I am quite certain they are not equivalent, but I have no way of checking which is correct. How should one approach a question like this? Perhaps a solution done on paper would be quicker, I don't mind.

Answer 1

Taking this step by step, apply the chain rule once to achieve:

$${\mathrm dy \over \mathrm dx} = -{4\over u^5}{\mathrm du \over \mathrm dx}\quad \quad u = x+\sqrt{4+\left(x^2-1\right)^3}$$

Finding ${\mathrm du \over \mathrm dx}$:

$${\mathrm du \over \mathrm dx} = \left(1+{1\over 2\sqrt v}{\mathrm dv \over \mathrm dx}\right)\quad \quad v = 4+\left(x^2-1\right)^3$$

Finding ${\mathrm dv \over \mathrm dx}$:

$${\mathrm dv \over \mathrm dx} = 3\left(x^2-1\right)^2\cdot 2x$$

And substituting back:

$$\begin{align} {\mathrm dy \over \mathrm dx} &= -{4\over \left(x+\sqrt{4+\left(x^2-1\right)^3}\right)^5}\cdot \left(1+{1\over 2\sqrt{4+\left(x^2-1\right)^3}} \cdot 3\left(x^2-1\right)^2\cdot 2x\right) \\ &= -{4\over \left(x+\sqrt{4+\left(x^2-1\right)^3}\right)^5}\cdot \left(1+{3x\left(x^2-1\right)^2\over \sqrt{4+\left(x^2-1\right)^3}}\right) \\ \end{align}$$

It looks like you forgot to consider the chain rule for $\left(x^2-1\right)^3$.

Answer 2

Let

$a=x+\sqrt{4+\left(x^2-1\right)^3}$

$b=4+\left(x^2-1\right)^3$

$c=(x^2+1)$

Then

$\frac{dy}{dx}=\frac{d}{dx}a^{-4}=-4a^{-5}\frac{d}{dx}a$

$\frac{d}{dx}a=1+\frac{d}{dx}b^{\frac{1}{2}}=1+\frac{1}{2}b^{-\frac{1}{2}}\frac{d}{dx}b$

$\frac{d}{dx}b=0+\frac{d}{dx}c^{3}=3c^{2}\frac{d}{dx}c$

$\frac{d}{dx}c=2x$

Therefore, $\frac{dy}{dx}=-4a^{-5}(1+\frac{1}{2}b^{-\frac{1}{2}}3c^{2}2x)=-4a^{-5}(1+3xc^{2}b^{-\frac{1}{2}})$

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