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.
So our function states y equals to ln of 10x plus e to the 3x.
First step, look at your question, look at it closely and think about it.
You can see here that you have ln of something and then you have something inside of it.
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.
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.
Notice that this inside part hasn't change at all. It's exactly the same.
So ln of that bracket goes to 1 over that bracket,
don't forget now we now multiply with the differential of the inside.
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.
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.
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.