How do you differentiate #y = tan^3(x^3)#?

1 Answer
Jim H
Jul 27, 2015

Use the chain rule twice.

Explanation:

It is important to remember the convention for powers of trigonometric functions:

#tan^3(x^3)# means "raise #x# to the third power, then find the tangent of that number and finally raise that tangent to the third power". In notation:

#(tan(x^3))^3#

The outermost function is the (second) cube.
And we know that #d/dx (u^3) = 3u^2 (du)/dx#
In this case #u = tan(x^3)#

So we get

#dy/dx = 3(tan(x^3))^2 * d/dx(tan(x^3))#

Now we need the derivative of #tan(x^3)#.

We'll use the chain rule again:

#d/dx(tanu) = sec^2u (du)/dx#

We get

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

#= sec^2(x^3)* 3x^2#

Putting it all together, we have:

#y = tan^3(x^3)#

#dy/dx = 3tan^2(x^3)*sec^2(x^3)*3x^2#

Which can be written:

#dy/dx = 9x^2tan^2(x^3)sec^2(x^3)#

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