# How to differentiate tany = ((3x-x^3)/(1-3x^2)) ?

## $\tan y = \left(\frac{3 x - {x}^{3}}{1 - 3 {x}^{2}}\right)$

Jun 13, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{{x}^{2} + 1}$

#### Explanation:

I assume we're differentiating with respect to $x$.

$\frac{d}{\mathrm{dx}} \tan y = \frac{d}{\mathrm{dx}} \left(\frac{3 x - {x}^{3}}{1 - 3 {x}^{2}}\right)$

On the left, we need to use the chain rule. On the right, we'll use the quotient rule.

${\sec}^{2} y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = \frac{\left(\frac{d}{\mathrm{dx}} \left(3 x - {x}^{3}\right)\right) \left(1 - 3 {x}^{2}\right) - \left(3 x - {x}^{3}\right) \left(\frac{d}{\mathrm{dx}} \left(1 - 3 {x}^{2}\right)\right)}{1 - 3 {x}^{2}} ^ 2$

Finding these derivatives and simplifying:

${\sec}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(3 - 3 {x}^{2}\right) \left(1 - 3 {x}^{2}\right) - \left(3 x - {x}^{3}\right) \left(- 6 x\right)}{1 - 3 {x}^{2}} ^ 2$

${\sec}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {x}^{4} + 6 {x}^{2} + 3}{1 - 3 {x}^{2}} ^ 2$

You may notice the simplification:

${\sec}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {\left({x}^{2} + 1\right)}^{2}}{1 - 3 {x}^{2}} ^ 2$

Now, we solve for the derivative:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sec} ^ 2 y \frac{3 {\left({x}^{2} + 1\right)}^{2}}{1 - 3 {x}^{2}} ^ 2$

It's weird to leave the answer in this form, but we have a way around it! Recall that ${\tan}^{2} y + 1 = {\sec}^{2} y$.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{{\tan}^{2} y + 1} \frac{3 {\left({x}^{2} + 1\right)}^{2}}{1 - 3 {x}^{2}} ^ 2$

And we know that $\tan y = \frac{3 x - {x}^{3}}{1 - 3 {x}^{2}}$:

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

Let's simplify that:

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

Cancel the ${\left(1 - 3 {x}^{2}\right)}^{2}$ terms and proceed by expanding ${\left(3 x - {x}^{3}\right)}^{2} + {\left(1 - 3 {x}^{2}\right)}^{2}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {\left({x}^{2} + 1\right)}^{2}}{{x}^{6} + 3 {x}^{4} + 3 {x}^{2} + 1}$

Which can be simplified:

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{{x}^{2} + 1}$

Jun 13, 2018

$\frac{3}{1 + {x}^{2}}$.

#### Explanation:

Recall that, $\tan 3 \theta = \frac{3 \tan \theta - {\tan}^{3} \theta}{1 - 3 {\tan}^{2} \theta}$.

So, if we subst. $x = \tan \theta$ [this we can do, because the range of

$\tan$ function is $\mathbb{R}$], then,

$\tan y = \tan \left(\frac{3 x - {x}^{3}}{1 - 3 {x}^{2}}\right)$,

$\Rightarrow y = {\tan}^{-} 1 \left(\frac{3 x - {x}^{3}}{1 - 3 {x}^{2}}\right)$,

$= {\tan}^{-} 1 \left\{\tan \left(\frac{3 \tan \theta - {\tan}^{3} \theta}{1 - 3 {\tan}^{2} \theta}\right)\right\}$,

$= {\tan}^{-} 1 \left(\tan 3 \theta\right)$.

$\Rightarrow y = 3 \theta = 3 {\tan}^{-} 1 x$.

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = 3 \cdot \frac{1}{1 + {x}^{2}} = \frac{3}{1 + {x}^{2}}$, as Respected mason m. has