Professor Dave here, let’s learn the chain rule.
We know how to take derivatives of polynomials in standard form, trigonometric functions,
as well as simple products and quotients of these functions.
But unfortunately, functions can get more complicated than this.
What if we have a trinomial raised to an exponent?
Or a trig function operating on a polynomial?
When we have multiple operations happening at once within a function, in order to take
its derivative, we will have to use something called the chain rule.
To see how this works, let’s look at this function, the square root of (x squared plus one).
None of the rules of differentiation that we have learned so far will help us with this.
That’s because this is actually a composite function.
To our input x, we are squaring and adding one, and then we are taking the square root
of all that, which as we recall, means raising it to the one half power.
So we could really say that f of x is (x to the one half) and g of x is (x squared plus
one), and if we call the composite function capital F of x, in trying to find the derivative
of capital F of x, we are actually trying to take the derivative of f of g of x.
The way we do this, according to the chain rule, is that we find f prime of g of x, and
then multiply it by g prime of x.
What that means is that moving from the outside in, we can take the derivative of the function
as though this exponent is the only operation, leaving the inside as it is.
But since the inside is another function, we have to multiply our result by the derivative
of what’s inside.
So looking at (x squared plus one) to the one half power, let’s apply the power rule.
Now we are used to using the power rule for positive integer exponents, but don’t worry,
we can do this with fractional exponents too, and it won’t be any different.
We just take the exponent and bring it to the front, and then subtract one from the
exponent, so we get one half times this parenthetical term, which is now raised to the negative
one half power.
If we recall our rules of exponents, raising something to a negative exponent is the same
as one over that thing raised to the positive version of that exponent, so this becomes
one over the quantity (x squared plus one) to the one half power, and we know that this
simply means the square root, so we can go back to expressing this as root (x squared
plus one), incorporate the one half by putting two in the denominator, and this is what we get.
But now we must remember, according to the chain rule, we have to multiply this by the
derivative of the function inside the parenthetical as well, and the derivative of (x squared
plus one) is two x, so let’s put two x up top, cancel out the two, and we are left with
x over root (x squared plus one).
So with the chain rule, when we have multiple functions happening at once, we differentiate
the outer function first, as it operates on the inner function and keeping the inner function
the same, and then we multiply by the derivative of the inner function.
Let’s try a few more for practice.
Say we want to get the derivative of the sine of x squared.
So we can clearly see that here there are two functions operating on x, the outer function
is sine, and the inner function is this exponent.
So let’s apply the chain rule.
The derivative of sine is cosine, so we get cosine x squared, but then we have to multiply
by the derivative of the inner function, x squared, which gives us two x, so we get two
x cosine x squared.
Just to see the reverse scenario, what if we have sine squared of x.
Let’s recall that this notation implies that the function sine of x is being squared,
so we can also express that like this, with sine x in parentheses, being raised to the
second power.
Now the situation is reversed, and the outer function involves squaring the input, while
the inner function is sine.
So let’s take the derivative of the outer function, which means pulling the two down
here to get two sine x.
But then we multiply by the derivative of the inner function, and the derivative of
sine is cosine, so we get two sine x cosine x.
Apart from good practice, these two examples should also prove to you that it is indeed
important the order in which we apply the chain rule, as we do get different answers.
It’s the derivative of the outer function as applied to the inner function, times the
derivative of the inner function.
Now let’s try some other examples.
Take the quantity (x to the fourth minus one) raised to the fiftieth power.
This huge exponent might seem daunting, but let’s just apply the chain rule.
The outer function is this exponent, so we take the fifty and put it down here, leave
what’s inside the parentheses just as it is, and reduce the exponent to forty nine.
This is no different than what we are used to.
But then we multiply by the derivative of the inner function, and that will be four
x cubed, so combining these terms, we get (two hundred x cubed) times the quantity (x
to the fourth minus one) raised to the forty-ninth power.
A little messy perhaps, but not difficult at all.
However, there are instances in which we will have to combine the chain rule with the power
rule or quotient rule.
Take for example (x minus one) over (x plus one) quantity squared.
Naturally we need the chain rule.
We take the derivative of the outer function, and we get two times this parenthetical.
But now, we must multiply by the derivative of the inner function, and to do that we need
the quotient rule.
Let’s save our place in this problem so that we can come back to it once we find this
other derivative, and set up our quotient.
That will be (x plus one) times one, minus (x minus one) times one, over x plus one quantity squared.
Simplifying the top, and being careful to distribute this negative sign, we get two
over x plus one quantity squared.
Now since this is the derivative of this quotient, we can plug this in back here, where we needed it.
We can simplify by combining the 2’s in the numerator to get four, and since the denominators
have the same term, let’s just make this (x plus one) quantity cubed.
And that’s pretty much all we can do.
Let’s try another.
(X plus three) quantity squared times (x squared minus four) quantity cubed.
This time we will have to use the product rule and the chain rule, but we will have
to use the product rule first.
This means the first function times the derivative of the second, plus the second function times
the derivative of the first.
But for each derivative, we must use the chain rule, so let’s expand these portions.
For the first derivative, we pull the three out front and change the exponent to a two,
remembering to then multiply by the derivative of what’s inside, which is two x.
For the other, we do the same thing, getting two times the quantity x plus three, and the
derivative of what’s inside is simply one in this case.
From here, most of the work is already done, so let’s just simplify a little bit until
we get two terms.
From here we could identify the greatest common factor of these two terms, and factor it out.
Then just distribute everything that’s left inside to get one term, and this is therefore
the most simplified way to express our answer.
Let’s do just one more example to illustrate how crucial the chain rule is.
What if we have sine of cosine of tangent of x.
This is a new situation, as we see three functions all operating at once.
But let’s not worry too much, let’s just apply the chain rule the same way we already
have, working from the outside in.
The outermost function is sine, so let’s not worry what sine is operating on, since
that won’t change.
Just think of it as being equal to U for the time being.
The derivative of sine is cosine, so we get cosine of U, or the cosine of what was originally
in there, so cosine of cosine of tangent x.
But then we have to multiply by the derivative of what was inside, and the derivative of
U, or the derivative of cosine of tangent of x, will itself require the application
of the chain rule.
So let’s just keep working our way through this, not touching any of the work we’ve
already done, but now evaluating this other derivative using the chain rule.
For this part, the outer function is cosine, and the derivative of cosine is negative sine,
so we get the negative sine of tangent x, but then we multiply by the derivative of
the inner function, so the derivative of tangent x, and if we remember our derivatives of trig
functions from the previous tutorial, that is equal to secant squared x.
So this is what we are left with.
All we can do is bring the negative sign to the front, and we get negative cosine of cosine
of tangent x, times sine of tangent x, times secant squared x.
So we can see that even when looking at three, or four, or literally any number of functions
operating all at once, there is nothing new to apply, it’s all about applying the chain
rule as many times as we need to.
Of course things can get pretty messy, but as long as we stay organized, using the right
notation and applying the chain rule properly, we should have no problem getting the right answer.
Let’s check comprehension.