Question

How do you find the derivative of `y=lne^x`?

Answer 1

`dy/dx=1`

Explanation:

We should know for this approach that `d/dxln(x)=1/x`.

Applying the chain rule to this derivative tells us that if we were to have a function `u` instead of just the variable `x` within the logarithm, we see that `d/dxln(u)=1/u*(du)/dx`.

So we see that `d/dxln(e^x)=1/e^x*d/dx(e^x)`

Since `d/dx(e^x)=e^x`:

`d/dxln(e^x)=1/e^x*e^x=1`

Answer 2

`dy/dx=1`

Explanation:

The logarithm function and exponential functions are inverse functions--they undo one another! This means that `log_a(a^x)=x` and `a^(log_a(x))=x`.

Recall that the function `ln(x)` is the logarithm with a base of `e`, that is, `ln(x)=log_e(x)`. Thus:

`y=ln(e^x)=log_e(e^x)=x`

So

`dy/dx=1`