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# Derivative of a^x proof

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Hey! I was recently trying to prove the derivative of a^x.

After trying it myself I ended up having to Google it because I don't seem to get it quite right. Here is what I tried to do:

$$y = {a^{x}}$$

$$ln(y) = x*ln(a)$$

$$\frac{d}{dx} ln(y) = \frac{d}{dx} x*ln(a)$$

Since ln(a) is a constant I just take the drivative of x which is one.

$$\frac{1}{y} = ln(a)$$

From above we see that y = a^x so I substitute it in there.

$$\frac{1}{a^{x}}= ln(a)$$

Rearranging gives me:

$$1= ln(a) * a^{x}$$

Here comes my problem, the result is supposed to be:

$$\frac{dy}{dx}= ln(a)*a^{x}$$

the dy/dx is supposed to come up on the left hand side as I take the derivative of ln(y) and get 1/y. I don't understand why this happens and I haven't seen any explanation behind for that specific step. Everything else is logical, just logarithmic/exponential properties. I would appreciate any help, thanks in advance!

Quazars  Nov 9, 2015

#1
+88775
+15

y = a^x   take the ln of both sides

lny = lna^x    and we can write

lny =  ln a^x    exponentiate both sides

e ^(ln y)   = e^(ln a^x)

y  = e^(ln a^x)

y = e^(x ln a)    take the derivative

y ' = lna * e^(x ln a)

y '  = lna * e^(ln a^x)

y ' = lna * a^x      and we can write

dy / dx   = (ln a) * a^x

CPhill  Nov 9, 2015
#1
+88775
+15

y = a^x   take the ln of both sides

lny = lna^x    and we can write

lny =  ln a^x    exponentiate both sides

e ^(ln y)   = e^(ln a^x)

y  = e^(ln a^x)

y = e^(x ln a)    take the derivative

y ' = lna * e^(x ln a)

y '  = lna * e^(ln a^x)

y ' = lna * a^x      and we can write

dy / dx   = (ln a) * a^x

CPhill  Nov 9, 2015
#2
+93289
+3

That is neat Chris :))

Melody  Nov 10, 2015
#3
+5

At the point where you have $$\frac{d}{dx}\ln(y),$$

you are required to differentiate a function of y wrt x.

To do this you should be using the function of a function rule, which, in this case, takes the form

$$\displaystyle \frac{df(y)}{dx}=\frac{df(y)}{dy}\times\frac{dy}{dx}.$$

Guest Nov 10, 2015