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

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

2 Answers
Jun 13, 2018

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)

Jun 13, 2018

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!