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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=ax

ln(y)=xln(a)

ddxln(y)=ddxxln(a)

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

1y=ln(a)

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

1ax=ln(a)

Rearranging gives me:

1=ln(a)ax

 

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

dydx=ln(a)ax

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!

 Nov 9, 2015

Best Answer 

 #1
avatar+130466 
+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

 

 

cool cool cool

 Nov 9, 2015
 #1
avatar+130466 
+15
Best Answer

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

 

 

cool cool cool

CPhill Nov 9, 2015
 #2
avatar+118696 
+3

That is neat Chris :))

 Nov 10, 2015
 #3
avatar
+5

At the point where you have ddxln(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

df(y)dx=df(y)dy×dydx.

 Nov 10, 2015

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