+267 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!
+128407 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
+128407 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
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
\(\displaystyle \frac{df(y)}{dx}=\frac{df(y)}{dy}\times\frac{dy}{dx}.\)
