How do you take the derivative of # 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)#

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))#