Differentiation of function x^x

Patjamet
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Just wondering if I have done this correctly?

Homework Statement


Differentiate:

x^{sin3x}

The Attempt at a Solution



x^{sin3x}=e^{sin(3x)ln(x)

Employing chain rule.

y=e^{u}
u=sin(3x)ln(x)

dy/dx = e^{u}\times(sin(3x)/x + 3ln(x)cos(3x))

Final Solution? =

dy/dx = e^{sin(3x)ln(x)}\times(sin(3x)/x + 3ln(x)cos(3x))
 
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EDIT: Made a mistake while entering your proposed answer, you're correct.
 
Last edited:
Patjamet said:
x^{sin3x}=e^{sin(3x)ln(x)}

The OPs result is equivalent to yours.
 
Thanks guys.

May I ask what the "OP" is?
 
Patjamet said:
Thanks guys.

May I ask what the "OP" is?

Original Post or Original Poster.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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