Differentiating $\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$

Question

Find the derivative with respect to $x$ of $$\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$$

The partial solution to this problem is given as follows: $y=\left(\frac{x}{\sqrt{1-x^2}}\right)$

Then: $\frac{dy}{dx} = (1-x^2)^{-3/2}$

Can you show the steps in arriving at this partial solution of $\frac{dy}{dx}$?

Product Rule with Chain Rule, Quotient Rule with Chain Rule, and I end up with additional terms. Notably, an $x^2$ multiplying the $(1-x^2)^{-3/2}$, plus an additional term of $(1-x^2)^{-1/2}$

Answer 1

Hint: $$\arctan{\left(\frac{x}{\sqrt{1-x^2}}\right)}=\arcsin{(x)}$$

Answer 2

Yes I think you are right.

$$\frac{dy}{dx}=(1-x^2)^{-\frac{1}{2}}+x(1-x^2)^{-\frac{3}{2}}(-2x)$$ $$=(1-x^2)^{-\frac{3}{2}}(1-x^2-2x^2)$$ $$=(1-x^2)^{-\frac{3}{2}}(1-3x^2)$$

The partial solution has to be wrong.

Answer 3

substitute $x=sin\theta$,then $y=arctan(\frac{x}{\sqrt{1-x^2}})$ becomes $y=\theta$ or $y=-\theta$ depending on the sign of $cos\theta$.

Then you can use the chain rule.So we have $\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$. Now i think you can complete it.

Answer 4

Continuing to work through the solution provided I have arrived at an answer. It is as follows:

$\frac{d}{dx} \frac{x}{\sqrt(1-x^2)}$

$\frac{d}{dx} x(1-x^2)^{-1/2}$

Using the Product Rule (u' * v) + (v' * u) where u = $x$ and v = $(1-x^2)^{-1/2}$ and combining with the Chain Rule for the derivation of v' yields the following:

$(1)*(1-x^2)^{-1/2} + (\frac{-1}{2}(1-x^2)^{-3/2}*(-2x)*(x)$

NOTE: The Chain Rule applied to v' gave us this: $\frac{-1}{2}(1-x^2)^{-3/2}*(-2x)$

Now, further simplifying ...

$(1-x^2)^{-1/2} + (\frac{2x^2}{2}(1-x^2)^{-3/2}$

$(1-x^2)^{-1/2} + (x^2(1-x^2)^{-3/2}$

The above was where I was stuck, but here is the breakthrough ...

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

Get a common denominator ...

$\frac{(1-x^2)^{-1/2}(1-x^2)^{3/2} + x^2}{(1-x^2)^{3/2}}$

$\frac{(1-x^2)^{2/2} + x^2}{(1-x^2)^{3/2}}$

$\frac{1-x^2+x^2}{(1-x^2)^{3/2}}$

$\frac{1}{(1-x^2)^{3/2}}$

$=(1-x^2)^{-3/2}$

NOTE: This is only the partial solution to the problem. For those interested in the full solution read on ...

Without showing this derivation, I know

$\frac{d}{dx} tan^{-1}(x) = \frac{1}{1+x^2}$

We also know from the original problem that $x = \frac{x}{\sqrt(1-x^2)}$

So now substitute $x = \frac{x}{\sqrt(1-x^2)}$ into derivation of $\frac{d}{dx} tan^{-1}(x) = \frac{1}{1+x^2}$

This yields

$\frac{d}{dx} tan^{-1}(x) = \frac{1}{1+x^2} = \frac{1}{1+\left(\frac{x}{\sqrt(1-x^2)} \right)^2}$

Square the term in the denominator

$=\frac{1}{1+\frac{x^2}{1-x^2}}$

Get a common denominator

$=\frac{1}{\frac{1-x^2+x^2}{1-x^2}}$

Simplify

$=1-x^2$

Now, finishing up ...

The derivative with respect to $x$ of $tan^-1\left(\frac{x}{\sqrt(1-x^2)}\right)$ requires the Chain Rule. This is:

$\frac{d}{dx}tan^{-1}\left(\frac{x}{\sqrt(1-x^2)}\right) * \frac{d}{dx}\frac{x}{\sqrt(1-x^2)}$

So, the derivative of the outside function $\frac{d}{dx}tan^{-1}\left(\frac{x}{\sqrt(1-x^2)}\right)$, or think of it just like the derivative of the $tan^{-1}(x)$ where we later substitute back in $x = \frac{x}{\sqrt(1-x^2)}$, times the derivative of the inside function $\frac{d}{dx}\frac{x}{\sqrt(1-x^2)}$. We have already found both derivations above. Let's substitute the derivations in and finish solving this problem.

By Chain Rule: $\frac{d}{dx}tan^{-1}\left(\frac{x}{\sqrt(1-x^2)}\right) = \frac{d}{dx}\left(tan^{-1}\left(\frac{x}{\sqrt(1-x^2)}\right)\right) * \frac{d}{dx}\left(\frac{x}{\sqrt(1-x^2)}\right)$

$=(1-x^2) * (1-x^2)^{-3/2}$

$=(1-x^2)^{2/2} * (1-x^2)^{-3/2}$

$=(1-x^2)^{-1/2}$

$=\frac{1}{\sqrt(1-x^2)}$