Your minds must be in Calculus overdrive by now.
There is so much that you know. We know how to differentiate from first principles.
We have looked at how to differentiate the easy way.
We have looked at three different methods of differentiation and the kind of functions.
You then move to differentiating trigonometry and you're now putting it all together.
We'll move on to the next section where we start to look at differentiation of exponential and logs.
We're going to introduce how to do it, so just the basic rules,
and then we're going to do some questions. We'll also do some proving of log rules,
so there's three really important rule of logs that will proof.
Now all of these is really important for understanding things like the pH scale in medicine,
so how acidic or alkaline something is. So once you'll be able to deal with log calculations,
you should be a lot of more confident with your calculations on the pH scale. So let's start.
The first thing, just defining what e of x which is the exponential
and log of x which will deal with in this case as ln of x and defining what the derivatives are.
The easiest exponentials to deal with is just e of x which is a naturally I can remember.
And the good thing is that the differential of e of x is simply e of x.
It doesn't change, so if we were to sketch the gradient of e of x
it would look exactly the same as the graph of e of x, it makes our life easier.
So e of x differentiates the e of x. Now, we're going to use a special log called ln of x.
Which is basically log base e. So the e that we can just see here,
ln of x is log of base e and we can shorten it to ln of x.
An ln of x differentiate to 1/x.
Now, this will derive later on when we do some integration, but for now,
we'll just go straight into some calculations and I will also give you three important rules of logs,
which you may have already seen in the pass,
but we're going to prove them so you understand them a little better
and again like I said, they'll be extremely helpful for you
when you use or apply log rules to your pH scales.
So using the rules that we've just looked at, let's do our first example.
We have y equals to 5e to the x, plus 7. Now, this is fairly straight forward question.
All we have to do is differentiate each term separately.
So if I was to do dy/dx, remember the 5 is just a constant,
so if it's multiplying with the function it doesn't change.
We now need to differentiate e of x, and if you look back you'll see the e of x differentiates to e of x.
So this doesn't change, and the plus 7 is a constant, so that just disappears.
So simply put the differential of 5e to the x plus 7, it's just 5e to the x.
Let's build up in difficulty, and look at our second example.
We now have y equals to 7e to the 5x minus 1, plus 3x.
Now you see this is a little bit different to the previous question,
because in the previous question you just have to differentiate e of x
and you know that e of x differentiates to e of x.
However, in this case you are now dealing with e of 5x minus 1.
What is that look like, if you really think about it?
You have e, not to the power of x but to the power of another expression
or to the power of another function. Think about it for a moment
and hopefully you will conclude that this is a function of a function.
So you have a function and outside function of e and an inside function of 5x minus 1.
So again, we have to use the chain rule here,
let's try it out and see if we can do this together.
So 7 is just a constant, doesn't bother us, that can stay.
Right, I'm just going to show you here on this side, 5x minus 1,
the outside function is e and it's e to the power of something.
Remember when we use the chain rule, the outside we differentiate the outside function first
and then we multiply it with the differential of the inside function.
But the good thing here or perhaps the unusual thing here,
is when we differentiate e to the power of anything as a whole, it doesn't change.
Because e to the x differentiate to e to the x.
So e to the 5x minus 1, when you look at as the outside function
to simply goes to e to the 5x minus 1.
And then don't forget we have to multiply this with the differential of the inside function.
So we can say that this is our inside function.
Differential of 5x minus 1 is 5 so the answer to that is just going to be 5e to the 5x minus 1.
So that is the differential of e to the 5x minus 1 that is 5e to the 5x minus 1.
And then 3x, we're experts at this, so 3x just differentiates to 3.
Let's just tidy this up, 7 times 5 gives you 35, e to the 5x minus 1,
plus 3 and that is your derivative of 7e to the power of 5x minus 1, plus 3x.