Differentiation of Exponentials and Logs: Exercise 3

00:01 Moving on to our next question, you can see here that you have an ln function, within I can also see an e function so let's just give this a go and break it down into one step at a time and then see what we get to.

00:14 So our function states y equals to ln of 10x plus e to the 3x.

00:20 First step, look at your question, look at it closely and think about it.

00:27 You can see here that you have ln of something and then you have something inside of it.

00:33 So let me just remind you that when you have a function y equals to ln of x the derivative of this dy/dx is 1/x, but remember that in this case we're dealing with x, so ln of x goes to 1/x, ln of something else, will go to 1 over something else.

00:51 So when we start to differentiate this dy/dx start ignore everything on the inside, so ignore what's in here. Just do ln of the brackets, we'll go to 1/10x plus e to the 3x.

01:06 Notice that this inside part hasn't change at all. It's exactly the same.

01:11 So ln of that bracket goes to 1 over that bracket, don't forget now we now multiply with the differential of the inside.

01:20 So the differential of the inside 10x differentiates to 10 and e to the 3x, so if I have e to the 3x, that will differentiate to, use the chain rule outside function, inside function.

01:35 So e to the 3x just stays as e to the 3x and then multiply it with the differential of this so that's 3e to the 3x.

01:44 So this becomes plus 3e to the 3x and all you have to do is just tidy this up, 10 plus 3e to the 3x all over 10x, plus e to the 3x.