Differentiation of exponents x

[cal.queen92]

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?

 

[Saladsamurai]

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.

 

[Mark44]

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.

= 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?

 

[cal.queen92]

That helps! Awsome! It worked out, thank you!