Question

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

`tany = ((3x-x^3)/(1-3x^2))`

Answer 1

`dy/dx=3/(x^2+1)`

Explanation:

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

`d/dxtany=d/dx((3x-x^3)/(1-3x^2))`

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

`sec^2y(dy/dx)=((d/dx(3x-x^3))(1-3x^2)-(3x-x^3)(d/dx(1-3x^2)))/(1-3x^2)^2`

Finding these derivatives and simplifying:

`sec^2ydy/dx=((3-3x^2)(1-3x^2)-(3x-x^3)(-6x))/(1-3x^2)^2`

`sec^2ydy/dx=(3x^4+6x^2+3)/(1-3x^2)^2`

You may notice the simplification:

`sec^2ydy/dx=(3(x^2+1)^2)/(1-3x^2)^2`

Now, we solve for the derivative:

`dy/dx=1/sec^2y(3(x^2+1)^2)/(1-3x^2)^2`

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

`dy/dx=1/(tan^2y+1)(3(x^2+1)^2)/(1-3x^2)^2`

And we know that `tany=(3x-x^3)/(1-3x^2)`:

`dy/dx=1/((3x-x^3)^2/(1-3x^2)^2+1)(3(x^2+1)^2)/(1-3x^2)^2`

Let's simplify that:

`dy/dx=(1-3x^2)^2/((3x-x^3)^2+(1-3x^2)^2)(3(x^2+1)^2)/(1-3x^2)^2`

Cancel the `(1-3x^2)^2` terms and proceed by expanding `(3x-x^3)^2+(1-3x^2)^2`:

`dy/dx=(3(x^2+1)^2)/(x^6+3x^4+3x^2+1)`

Which can be simplified:

`dy/dx=(3(x^2+1)^2)/(x^2+1)^3`

`dy/dx=3/(x^2+1)`

Answer 2

`3/(1+x^2)`.

Explanation:

Recall that, `tan3theta=(3tantheta-tan^3theta)/(1-3tan^2theta)`.

So, if we subst. `x=tantheta` [this we can do, because the range of

`tan` function is `RR`], then,

`tany=tan((3x-x^3)/(1-3x^2))`,

`rArr y=tan^-1((3x-x^3)/(1-3x^2))`,

`=tan^-1{tan((3tantheta-tan^3theta)/(1-3tan^2theta))}`,

`=tan^-1(tan3theta)`.

`rArr y=3theta=3tan^-1x`.

`:. dy/dx=3*1/(1+x^2)=3/(1+x^2)`, as Respected mason m. has

readily derived!