Differentiation of exponents x

cal.queen92
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Homework Statement



If f(x) = 2sin(x) + (4x)^x find ---> f ' (3)



Homework Equations



d/dx (e^u) = e^u du/dx



The Attempt at a Solution



Here is my attempt:

f(x) = 2sin(x) + e^(ln(4x)^x) = 2sin(x) + e^(xln4x) = d/dx (2sin(x)) + d/dx (e^(xln4x))

= 2cos(x) + e^(xln4x) d/dx (xln4x) = 2cos(x) + e^(ln(4x)^x)*(4x+(x/4x))

= 2cos(x) + ((4x)^x) * (4x+(1/4)) = (unsimplified) --> (2cos(3) + (12)^3)(12+(1/4)

But this answer is incorrect. What am i doing wrong?
 
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cal.queen92 said:

Homework Statement



If f(x) = 2sin(x) + (4x)^x find ---> f ' (3)



Homework Equations



d/dx (e^u) = e^u du/dx



The Attempt at a Solution



Here is my attempt:

f(x) = 2sin(x) + e^(ln(4x)^x) = 2sin(x) + e^(xln4x) = d/dx (2sin(x)) + d/dx (e^(xln4x))

= 2cos(x) + e^(xln4x) d/dx (xln4x) = 2cos(x) + e^(ln(4x)^x)*(4x+(x/4x))

= 2cos(x) + ((4x)^x) * (4x+(1/4)) = (unsimplified) --> (2cos(3) + (12)^3)(12+(1/4)

But this answer is incorrect. What am i doing wrong?

I believe your error lies in the bold parts. Bold parts should be equal. But \frac{d}{dx}(xln4x)\ne (4x+(x/4x). You should have a ln(4x) factor in there somewhere.
 
Last edited:
cal.queen92 said:

Homework Statement



If f(x) = 2sin(x) + (4x)^x find ---> f ' (3)



Homework Equations



d/dx (e^u) = e^u du/dx



The Attempt at a Solution



Here is my attempt:

f(x) = 2sin(x) + e^(ln(4x)^x) = 2sin(x) + e^(xln4x) = d/dx (2sin(x)) + d/dx (e^(xln4x))

= 2cos(x) + e^(xln4x) d/dx (xln4x) = 2cos(x) + e^(ln(4x)^x)*(4x+(x/4x))
Mistake at the end, above. When you differentiate x ln4x, you have to use the product rule and the chain rule.

d/dx(x ln(4x)) = ln(4x) + x * (1/(4x)) * 4 = ln(4x) + 1.
cal.queen92 said:
= 2cos(x) + ((4x)^x) * (4x+(1/4)) = (unsimplified) --> (2cos(3) + (12)^3)(12+(1/4)

But this answer is incorrect. What am i doing wrong?
 
That helps! Awsome! It worked out, thank you!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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