How do you find the derivative of f(x)= 4sinx+4x^x?

1 Answer
Jun 29, 2016

Differentiating the 4sinx is significantly easier than the 4x^x.

If f(x)=4sinx+4x^x, f'(x)=d/dx(4sinx)+d/dx(4x^x)
f'(x)=4d/dx(sinx)+4d/dx(x^x)

d/dx(sinx)=cosx (important one to learn!)
d/dx(x^x) can not be computed using the generalised power rule for differentiation, such that: d/dx(x^x)!=x(x^(x-1)) since x is not a constant.

We can compute it however, by using the following rule:
f^g=e^(glogf), where f denotes f(x), g denotes g(x) and log is the natural logarithm.
By extension: x^x=e^(xlogx) and therefore:
d/dx(x^x)=d/dx(e^(xlogx))

The whole point of this is that it's possible to take the derivative of this function easily. This is because the derivative of e^x is just e^x, so we only need to multiply through by the derivative of what's in the power (as given by the chain rule).
Therefore d/dx(e^(xlogx))=e^(xlogx)(d/dx(xlogx))
But there's no point leaving the result like that since we can just go back to x^x from e^(xlogx).
Also, d/dx(xlogx)=1+logx

So finally, f'(x)=4cosx+4x^x(1+logx)