Hey @Hartej_Dhiman,

I really appreciate your post. It’s something even I got confused in when I was going over the content myself. Let us start from the foundation. First, we will borrow the result from the lecture video “The inverse function and it’s derivative”, i.e.,

g'(y) = \frac{1}{f'(x)}, where \\
g(x) = f^{-1}(x), and \\
y = f(x)

If you are confused with this much, then I recommend you to watch this lecture video once again, since I have just stated the result from the lecture video. Now, let’s move on to understand how we can obtain the derivative of `log`

from this property. In the video “The derivative of log(x)”, it was established that `log(x)`

is the inverse of `e^x`

, i.e., `log(e^x) = x`

.

So, let’s say that `g(y) = log(y)`

and `f(x) = e^x`

, where `f`

and `g`

are inverse of each other. We invoke the result from before:

g'(y) = \frac{1}{f'(x)}, where \\
f'(x) = \frac{d}{dx} e^x = e^x, \\
=> g'(y) = \frac{1}{e^x}

Now, note that from the result, we also have y = f(x) = e^x. So, we substitute `e^x`

with `y`

on the RHS on the equation, so that we can get an expression for g'(y) in terms of `y`

. So, we have:

g'(y) = 1 / y, or \\
\frac{d}{dy} log(y) = \frac{1}{y}

Voila, we have the derivative of `log(y)`

or `log(x)`

Now, let’s go ahead, and break down your thoughts on this as well.

I guess this is very easy to understand. Since we have y = f(x), it means, we can rewrite this as x = f^{-1}(y), and now, we have simply substituted x with f^{-1}(y), in the above equation.

You are correct on this.

You are mistaken on this. One is not supposed to see any equality between these 2. The result states that \frac{1}{e^x} is equal to the derivative of log(y), as I have mentioned above.

Let us know if you still have any confusion on this, and we will be more than happy to help you out.

Cheers,

Elemento