Is (xe^(xln4))/4 equivalent to (4^x)ln4?

[whateva]

On my exam, we had to find the derivative of 4^x. This is what I did
Y=4^x
lny=xln4
y=e^xln4
and then finding the derivative for that I got, (xe^(xln4))/4
My professor said that it was wrong and even after I told her what I did to get the answer. She told me the answer was (4^x)ln4 . Which I know it is but I think this is still equivalent to my answer. Was I right? Regardless I still don't get the point :(

 

[Dr Transport]

take the derivative of $\ln(y)$ and then substitute $y$ back into the result to get the professors answer...

 

[Mastermind01]

y=e^xln4
and then finding the derivative for that I got, (xe^(xln4))/4

How did you get that as the derivative?

 

[whateva]

I did the chain rule, so I got x*1/4*e^xln4 . Which I now realize is wrong, it should've been x*1/4+ln4*e^xln4. But was I right with the y=e^xln4?

 

[Mastermind01]

I did the chain rule, so I got x*1/4*e^xln4 . Which I now realize is wrong, it should've been x*1/4+ln4*e^xln4. But was I right with the y=e^xln4?

You're still doing it wrong but yes $y=e^{xln4}$ is correct

 

[Dr Transport]

$$y = 4^x$$

$$ \ln(y) = \ln(4^x) = x\ln(4)$$

$$\frac{dy}{y} = dx \ln(4)$$

$$\frac{dy}{dx} = y \ln(4) = x^4 \ln(4)$$

proves the instructors answer and no need in taking exponentials...

 

[Anora]

You mean to say that the final answer is:
dy/dx = yln(4) = (4^x)ln(4)
Right?

 

[whateva]

Also, why is derivative of ln(4) not evaluated as 1/4?

 

[haushofer]

Because the derivative of a constant is zero.

 

[Dr Transport]

You mean to say that the final answer is:
dy/dx = yln(4) = (4^x)ln(4)
Right?

yes, typo...