Patjamet
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Just wondering if I have done this correctly?
Differentiate:
x^{sin3x}
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))
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))