# I Derivative of 4^x

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1. Apr 25, 2017

### 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 :(

2. Apr 25, 2017

### Dr Transport

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

3. Apr 25, 2017

### Mastermind01

How did you get that as the derivative?

4. Apr 25, 2017

### 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?

5. Apr 25, 2017

### Mastermind01

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

6. Apr 26, 2017

### 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.....

7. Apr 27, 2017

### Anora

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

8. Apr 27, 2017

### whateva

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

9. Apr 28, 2017

### haushofer

Because the derivative of a constant is zero.

10. Apr 28, 2017

### Dr Transport

yes, typo......