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

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

##### 2 Answers

#### Explanation:

I assume we're differentiating with respect to

#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

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

And we know that

#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

#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)#

#### Explanation:

Recall that,

So, if we subst. **range** of

**Respected mason m.** has

**readily derived!**