If f (x) = 4sin(x)+3(x^x), find f'(2).

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If f (x) = 4sin(x)+3(x^x), find f'(2).

Mathematics
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Find the derivative of f'(x) and then plug 2 into it.
i have difficulty on finding the f'(x)
The challenge is 3x^x. (1) Set y=3x^x. (2) ln y=xln3x. (3)Solve this as implicit differentiation with the left side giving you derivative of y and chain rule dy/dx. Solve for dy/dx. Move y quantity to right side of eq. Plug in value of y from (1). I am being evasive so you work it out yourself.

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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...
Show us what you got. Some times answers written in different forms meaning the same thing. Master's students write these answer things; they have no sense of humor.
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))...
counld you show me how to get this part (1+ln(x))? coz i got this part wrong and i didnt know how to do this. thanks!
y=3x^x ln(y) = ln(3x^x) * Note where you made a mistake, you can't drop the ^x in front of the log until you take care of the three in front of it, first. ln(y) = ln(3) + ln(x^x) ln(y) = ln(3) x + xln(x) Differentiate... y / y' = ln x + 1 y = 3x^x (lnx + 1)
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)
As far as I can tell you got the calculus right. The number is off two units one way or another. Calculators are very specific and it easy to send wrong info when calculating ln and cos. I think your teacher would give you credit. Then important lesson here is learning how to manipulate the variable when it is in the exponent.
The process was correct, but he needs to make sure he's using the logarithm rules correctly... ln(a*b^x) != x ln(a * b). Need to split it into ln a + ln (b^x) first.
how about differntiating x^x?
i thought its (x^x) (lnx)?
You would get the same thing, except for where you multiple x back in, at the end.. Think about it. All the step did where you split the log from ln (ax^x) into ln a + ln(x^x) was produce a constant which disappears in the differentiation. The only part that would produce something with the a in it, would be where you multiple each side by y, to isolate y' in the last step...
Yeah, a possible mistake. In step (2) try is 3xlnx.
multiply y back in* tired

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