Help with a calculus problem?

Question

Let f(x)=2sin(x)+3(x^x)
then f'(3)=

what I got was 2cos3 + 3(3^3)(1+log3) but its wrong

Answer 1

To differentiate x^x:

set y = x^x

then ln y = ln x^x = x ln x.
=> y' / y = ln x + x(1/x) = ln x + 1, so y' = x^x(ln x + 1).

Therefore f'(x) = 2 cos(x) + 3(x^x)(ln x + 1), so
f'(3) = 2cos 3 + 3(3^3)(ln 3 + 1).

To evaluate this, make sure your calculator is in radians, not degrees.

Answer 2

3(ln(3)+1)*3^3+2cos(3)
which is 168.008

Answer 3

f(x)=2sin(x)+3x^x
y=x^x
ln y = x ln x
y=e^(xlnx)
y'=(lnx+1)(e^(xlnx))=(lnx+1)(x^x)
f'(x)=2cos(x) + 3(lnx+1)(x^x)
f'(3)=168.0076104