How do you take the derivative of $y = {tan}^{2} (x^{3})$ $y=tan^2(x^3)$ ?

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How do you take the derivative of $y = {tan}^{2} (x^{3})$ $y=tan^2(x^3)$ ?

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Answer 1 · 2015-08-21T17:07:40

$y^' = 6x^2 * tan(x^3) * sec^2(x^3)$

Explanation:

You can differentiate this function by using the chain rule twice, once for $u^2$ , with $u = tan(x^3)$ , and once more for $tan(t)$ , with $t = x^3$ .

This will get you

$d/dx(y) = d/(du)u^2 * d/dx(u)$

$y^' = 2u * d/dxtan(x^3)$

The derivative of $tanx^3$ will be equal to

$d/dx(tant) = d/(dt)tan(t) * d/(dx)(t)$

$d/dx(tant) = sec^2t * d/dx(x^3)$

$d/dx(tan(x^3)) = sec^2(x^3) * 3x^2$

Your target derivative will thus be equal to

$y^' = 2 * tan(x^3) * 3x^2 * sec^2(x^3)$

$y^' = color(green)(6x^2 * tan(x^3) * sec^2(x^3))$

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How do you take the derivative of $y = {tan}^{2} (x^{3})$ $y=tan^2(x^3)$ ?

1 Answer

Explanation:

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How do you take the derivative of $y = {tan}^{2} (x^{3})$ $y=tan^2(x^3)$ ?