"Simple" differentiation question

[renedox]

There is a question which is as follows:

(4^x)/(ln4)

I have tried many times to get the answer but with no success, would someone be so kind as to show me how to do it?

The answer is:

4^x

My working is as follows:

f(x) = (4^x)/(ln4)
F'(x) = ((ln 4 * ln 4 * 4^x) - (4^x - (1/4))) / ((ln 4)^2)
f'(x) = (4^x) - ((4^x * (1/4)) / ((ln 4)^2))

...help?

[KLscilevothma]

F'(x) = ((ln 4 * ln 4 * 4^x) - (4^x - (1/4))) / ((ln 4)^2)
f'(x) = (4^x) - ((4^x * (1/4)) / ((ln 4)^2))
You cannot use quotient rule here.

Let y = 4x

take ln on both sides

ln y = x ln 4

Take the first derivatives on both sides

y'/y = ln 4

y' = y ln 4 = 4x ln 4

So,

F(x) = (4^x)/(ln4)

F'(x) = (1/ln 4) * y' (since the denominator is a constant) = 4^x

[Hurkyl]

Well, you can use the quotient rule here if you remember that the derivative of a constant (like ln 4) is zero.

(of course, you should just factor the constant out like you suggested)

[renedox]

Ahh yip, got it. Thanks guys :D