How do you differentiate #[x^(sin3x)^2]#?

1 Answer
May 27, 2018

Answer:

#x^(sin^2 3x)sin3x((sin3x)/x + 6 cos3xlogx)#

Explanation:

A strange function this one - were you trying to solve some particular problem when it arose, or did you just try to compile something tricky?

There's a potential ambiguity in the question; is it #(x^(sin3x))^2# or is it #x^(sin^2 3x)# ? The rule for evaluating towers of powers is to work from right to left (else one could have written the expression more simply), which suggests the latter. Is this what was meant? I'll supply both.

Answer 1: #x^(sin^2(3x))#

We have an x both in the base and the exponent of the expression, so we use the same technique to differentiate it as is used for #x^x#; working logarithmically.

Let #y=x^(sin^2 3x)#. Then #log(y)=sin^2 3x logx# and so #1/y dy/dx=(sin^2 3x)/x + 3*2sin3xcos3xlogx#. Thus

#dy/dx = x^(sin^2 3x)sin3x((sin3x)/x + 6 cos3xlogx)#.

Answer 2: #(x^(sin3x))^2#

Let #y=(x^(sin3x))^2#. Then #dy/dx=2x^(sin3x)d/dx(x^(sin3x))#.
Let #z=x^(sin3x)#, so #dy/dx=2x^(sin3x)dz/dx#.

As in the previous answer, proceed logarithmically:
#logz = sin3x logx#, so #1/z dz/dx = (sin3x)/x + 3cos3xlogx#, giving #dz/dx = x^(sin3x)((sin3x)/x + 3cos3xlogx)#

Plugging this back in, we see

#dy/dx=2x^(2sin3x) ((sin3x)/x + 3cos3xlogx)#