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Find the derivative of f'(x) and then plug 2 into it.

i have difficulty on finding the f'(x)

y=3x^x
lny=xln3x
y'/y=1*ln3x+(3/3x)*x
y'=(ln3x+1)3x^x
is that it?

Good job. I think you got it.

but i still got a wrong answer...

f'(x)=4(cosx)+(ln3x+1)*3x^x
f'(2)=4(cos2)+(ln6+1)*12
correct answer was 18.6531788205308

The derivative is going to be f'(x) = 4cos(x) + 3x^x(1+ln(x))...

Sorry, ln(y) = ln(3) x + xln(x)...
Should be... ln(y) = ln(3) + xln(x)...

i see! so we need to apply the product rule too when we take the ln?

Yeah, because you have a function of x (just x itself) times another function of x (being ln x)

how about differntiating x^x?

i thought its (x^x) (lnx)?

Yeah, a possible mistake. In step (2) try is 3xlnx.

multiply y back in* tired