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

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

$tan y = (\frac{3 x - x^{3}}{1 - 3 x^{2}})$ $tany = ((3x-x^3)/(1-3x^2))$

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Answer 1 · 2018-06-13T16:43:50

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

Explanation:

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

$\frac{d}{d x} tan y = \frac{d}{d x} (\frac{3 x - x^{3}}{1 - 3 x^{2}})$ $d/dxtany=d/dx((3x-x^3)/(1-3x^2))$

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

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

Finding these derivatives and simplifying:

${sec}^{2} y \frac{d y}{d x} = \frac{(3 - 3 x^{2}) (1 - 3 x^{2}) - (3 x - x^{3}) (- 6 x)}{{(1 - 3 x^{2})}^{2}}$ $sec^2ydy/dx=((3-3x^2)(1-3x^2)-(3x-x^3)(-6x))/(1-3x^2)^2$

${sec}^{2} y \frac{d y}{d x} = \frac{3 x^{4} + 6 x^{2} + 3}{{(1 - 3 x^{2})}^{2}}$ $sec^2ydy/dx=(3x^4+6x^2+3)/(1-3x^2)^2$

You may notice the simplification:

${sec}^{2} y \frac{d y}{d x} = \frac{3 {(x^{2} + 1)}^{2}}{{(1 - 3 x^{2})}^{2}}$ $sec^2ydy/dx=(3(x^2+1)^2)/(1-3x^2)^2$

Now, we solve for the derivative:

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

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

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

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

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

Let's simplify that:

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

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

$\frac{d y}{d x} = \frac{3 {(x^{2} + 1)}^{2}}{x^{6} + 3 x^{4} + 3 x^{2} + 1}$ $dy/dx=(3(x^2+1)^2)/(x^6+3x^4+3x^2+1)$

Which can be simplified:

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

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

Answer 2 · 2018-06-13T17:24:08

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

Explanation:

Recall that, $tan 3 θ = \frac{3 tan θ - {tan}^{3} θ}{1 - 3 {tan}^{2} θ}$ $tan3theta=(3tantheta-tan^3theta)/(1-3tan^2theta)$ .

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

$tan$ $tan$ 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!

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

$tan y = (\frac{3 x - x^{3}}{1 - 3 x^{2}})$ $tany = ((3x-x^3)/(1-3x^2))$

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Explanation:

Explanation:

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

tany = ((3x-x^3)/(1-3x^2)) tany=(3x−x31−3x2)tany = ((3x-x^3)/(1-3x^2))

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Explanation:

Explanation:

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$tan y = (\frac{3 x - x^{3}}{1 - 3 x^{2}})$ $tany = ((3x-x^3)/(1-3x^2))$