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In is video, we will learn how to differentiate composite functions by applying the chain rule.
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We will see how to apply this to simple functions initially.
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And then, weβll consider more complex functions such as trigonometric and reciprocal trigonometric functions.
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Firstly, a reminder of what composite functions are.
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They are essentially functions of a function.
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Suppose we have two functions, π of π₯ equals two π₯ plus five and π of π₯ equals π₯ cubed.
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The composite functions π of π of π₯ and π of π of π₯ are what we get if we composed these two functions in either order.
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We apply one, and then we apply the other.
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π of π of π₯ means we applied π first, giving π₯ cubed.
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And then, we take this as our input for the function π, which will give two π₯ cubed plus five.
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π of π of π₯, however, is the composite function we would get if we apply π first to give two π₯ plus five and then take this as our input to the function π, which would give two π₯ plus five all cubed.
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And if we distribute the parentheses and simplify, this gives eight π₯ cubed plus 60π₯ squared plus 150π₯ plus 125.
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So, we know how to compose functions.
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But what about finding their derivatives?
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Well, in this case, if youβre asked to find the derivatives of either π of π of π₯ or π of π of π₯, it wouldnβt be too bad.
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Because we could compose the functions first, manipulate them algebraically, and then differentiate the resulting polynomial.
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But suppose, instead, the power of π₯ in the function π of π₯ had been 10 or 20 rather than just three, it would be extremely time consuming and tedious to distribute all these parentheses in order to give a polynomial.
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So, it would be much more helpful for us to have a rule that allows us to differentiate a composite function.
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And indeed, there is one.
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Itβs called the chain rule.
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Weβll illustrate the chain rule by first finding the derivative of the composite function π of π of π₯.
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So, weβve defined the functions π of π₯ and π of π₯ to be two π₯ plus five and π₯ cubed, respectively.
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And we saw that the composite function π of π of π₯ was two π₯ plus five all cubed, which simplifies to eight π₯ cubed plus 60π₯ squared plus 150π₯ plus 125.
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Now letβs consider finding the derivative of this function.
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To do so, we need to recall the power rule, which tells us that the derivative with respect to π₯ of π, thatβs a constant, multiplied by π₯ to the power of π is πππ₯ to the power of π minus one.
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And we recall also that in order to find the derivative of a sum or difference, we can just differentiate each term separately and then add them together.
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So, differentiating π of π of π₯ then gives the derivative π of π of π₯ prime, which is 24π₯ squared plus 120π₯ plus 150.
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Remember, the derivative of a constant is just zero.
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So, when we differentiate that term of plus 125, it just gives zero.
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Now letβs see if we can manipulate this derivative to see if we can identify any relationship with the derivatives of π and π individually.
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Weβll first take out a common factor of six to give six multiplied by four π₯ squared plus 20π₯ plus 25.
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We may then notice that four π₯ squared plus 20π₯ plus 25 is actually a perfect square.
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Itβs equal to two π₯ plus five all squared.
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And as two π₯ plus five is our expression for π of π₯, this is actually equal to π of π₯ squared.
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But what about that six?
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Well, six is equal to two times three.
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So, we can write this whole derivative as two times three times π of π₯ all squared.
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But how does this help?
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Well, to see this, we need to find the derivatives of π and π.
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Applying the power rule, we see that π prime of π₯ is equal to two and π prime of π₯ is equal to three π₯ squared.
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So, that two in our derivative of the composite function is the same as π prime of π₯.
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Now three times π of π₯ all squared is actually the derivative of π evaluated at π of π₯.
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π prime of π₯ is three π₯ squared.
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So, π prime of π of π₯ is three π of π₯ squared.
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So, what have we found?
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Well, for this example, we found that the derivative of π of π of π₯ is equal to the derivative of π, thatβs the derivative of the inner function, multiplied by the derivative of π, thatβs the outer function, with the inner function still inside.
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Now this is an illustration of the chain rule.
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Itβs not a proof but that is beyond the scope of what weβre going to look at in this video.
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So, the chain rule then, it tells us that the derivative of the composite function π of π of π₯ is equal to π prime of π₯ multiplied by π prime of π of π₯.
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We can also express the chain rule using Leibnizβs notation.
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If π¦ is equal to π of π of π₯, and we let π’ equal π of π₯ so that π¦ becomes π of π’, a function of π’, then dπ¦ by dπ₯ is equal to dπ¦ by dπ’ multiplied by dπ’ by dπ₯.
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This may look quite complicated, but itβs actually a relatively straightforward process, as weβll see in our examples.
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Leibnizβs notation is really helpful because it makes the chain rule a little bit more intuitive.
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Remember that finding derivatives is all about small changes in π₯.
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So, letβs allow Ξπ’ to represent a small change in π’ as a result of a small change in π₯.
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In order to find the derivative of π¦ with respect to π₯ dπ¦ by dπ₯, we consider the difference quotient Ξπ¦ by Ξπ₯.
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We see that by multiplying both the numerator and denominator by Ξπ’, which must be nonzero, and then reordering the terms, we get Ξπ¦ over Ξπ’ multiplied by Ξπ’ by Ξπ₯.
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As Ξπ₯ tends to zero, so will both Ξπ’ and Ξπ¦, giving dπ¦ by dπ₯ equals dπ¦ by dπ’ multiplied by dπ’ by dπ₯.
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That is the chain rule.
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The chain rule allows us to differentiate a wide class of complex functions.
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Letβs look at some examples.
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Find the first derivative of the function π¦ equals five π₯ squared minus six to the power of six.
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Now we see that this is an example of a composite function.
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If we consider the first function to be five π₯ squared minus six and the second to be π₯ to the power of six.
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We take five π₯ squared minus six as the input to our second function, giving five π₯ squared minus six all to the power of six.
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As this is a composite function, we can apply the chain rule.
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The chain rule tells us that if π¦ is a function of π’ and π’ is a function of π₯, then dπ¦ by dπ₯ is equal to dπ¦ by dπ’ multiplied by dπ’ by dπ₯.
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So, we need to decide how weβre going to define the function π’.
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Well, we take π’ to be our first function.
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Itβs the part inside the parentheses, π’ equals five π₯ squared minus six.
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π¦, therefore, becomes a function of π’.
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π¦ equals π’ to the sixth, and π’ is a function of π₯.
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We need to find both dπ¦ by dπ’ and dπ’ by dπ₯, which we can do by applying the power rule.
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In the case of the dπ¦ by dπ’, we just need to think of all the π₯βs in the power rule as being π’βs.
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We have then that dπ¦ by dπ’ is equal to six π’ to the five, and dπ’ by dπ₯ is equal to 10π₯.
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We write down the chain rule and then make the relevant substitutions, giving dπ¦ by dπ₯ is equal to six π’ to the five multiplied by 10π₯.
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Now hereβs a really important point.
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That derivative of π¦ with respect to π₯ must be in terms of π₯, and, at the moment, we still have the variable π’ involved.
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So, we must make sure that we reverse our substitution.
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π’ is equal to five π₯ squared minus six, so we have six multiplied by five π₯ squared minus six to the power of five multiplied by 10π₯.
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Simplifying then, we have that the first derivative of the function π¦ equals five π₯ squared minus six to the power of six is 60π₯ multiplied by five π₯ squared minus six to the power of five.
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Now this illustrates a really powerful application of the chain rule, in fact, a general rule for finding the derivative of a bracket raised to a power.
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If we express the derivative as 10π₯ multiplied by six multiplied by five π₯ squared minus six to the power of five, then we see what we have is the derivative of the bracket, or the derivative of whatβs inside the parentheses, thatβs 10π₯, multiplied by the original power, six, multiplied by that bracket with the power reduced by one from what it was originally.
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This gives us the chain rule extension to the power rule.
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This tells us that if we have a function π of π₯ raised to a power, then the derivative is equal to π prime of π₯, thatβs the derivative of whatβs inside the parentheses, multiplied by π and then multiplied by π of π₯ with the power reduced by one, π of π₯ to the π minus one.
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This is particularly useful if we have really high powers.
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So, letβs see how we can apply this rule to another example.
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Determine the derivative of π¦ equals negative two π₯ squared minus three π₯ plus four to the power of 55.
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Now this is where we really see the importance of the chain rule.
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When we have an exponent as high as 55, we certainly donβt want to attempt to distribute all the parentheses.
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Instead, weβre going to use the chain rule extension of the power rule, which tells us that the derivative of π of π₯ to the π is π prime of π₯ multiplied by π multiplied by π of π₯ to the π minus one.
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So, π of π₯ will be that function inside the parentheses, negative two π₯ squared minus three π₯ plus four.
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We can apply the power rule to differentiate π of π₯, giving negative four π₯ minus three.
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Now we can work out dπ¦ by dπ₯.
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Itβs equal to π prime of π₯, thatβs negative four π₯ minus three, multiplied by π, thatβs 55, multiplied by π of π₯ to the power of π minus one, thatβs negative two π₯ squared minus three π₯ plus four to the power of 54.
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Thereβs no need to expand the parentheses.
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So, weβve found that dπ¦ by dπ₯ is equal to 55 multiplied by negative four π₯ minus three multiplied by negative two π₯ squared minus three π₯ plus four to the power of 54.
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And weβve done this by applying the chain rule extension to the power rule.
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We can also apply the chain rule more than once within the same problem.
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So, letβs consider an example of this.
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Find the first derivative of the function π¦ equals the square root of eight π₯ minus sin of nine π₯ to the power of eight.
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Here we have π¦ is equal to the square root of another function, so we have a composite function.
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Weβre, therefore, going to apply the chain rule.
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Weβre going to define π’ to be the function underneath the square root, so π’ is equal to eight π₯ minus sin of nine π₯ to the power of eight.
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Then, π¦ is equal to the square root of π’, which we can express using index notation as π’ to the power of one-half.
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The chain rule tells us that dπ¦ by dπ₯ is equal to dπ¦ by dπ’ multiplied by dπ’ by dπ₯.
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So, we need to find each of these derivatives.
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dπ¦ by dπ’ is relatively straightforward.
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Using the power rule, we get one-half π’ to the power of negative one-half.
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For dπ’ by dπ₯, the derivative of eight π₯ is just eight.
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But what about the derivative of sin of nine π₯ to the power of eight?
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We actually need to apply the chain rule again.
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We can let π equal this function.
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And we can change the notation a little to write it as sin nine π₯ to the power of eight.
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Itβs an equivalent notation, but it might it make it a little clearer how weβre going to find the derivative.
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We recall the chain rule extension to the power rule, which told us that if we had a function π of π₯ raised to a power π, then its derivative was π prime of π₯ multiplied by π multiplied by π of π₯ to the power of π minus one.
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Here, we have a function, sin of nine π₯ raised to a power eight, so we can apply the chain rule extension to the power rule.
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We need to recall one more rule which is that the derivative with respect to π₯ of sin ππ₯ is π cos ππ₯.
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So, we begin.
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The derivative of the part inside the parentheses is nine cos nine π₯.
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Then, we multiply by the power eight.
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And then, we have the function inside the parentheses written out again, but with the power reduced by one.
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Simplifying gives 72 cos nine π₯ sin to the power of seven nine π₯.
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So, now that we found both dπ¦ by dπ’ and dπ’ by dπ₯, we can substitute into the chain rule.
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We have then that dπ¦ by dπ₯ is equal to a half π’ to power of negative a half multiplied by eight minus 72 cos nine π₯ sin nine π₯ to the power of seven.
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We must also remember to replace π’ in terms of π₯.
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So, π’ is equal to eight π₯ minus sin of nine π₯ to the power of eight.
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Weβll also simplify the fractions.
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Dividing by that denominator of two leaves coefficients of four and 36 in the numerator.
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And we recall also that π’ to the power of negative a half is equal to one over root π’.
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So, our derivative dπ¦ by dπ₯ simplifies to four minus 36 cos nine π₯ sin nine π₯ to the power of seven all over the square root of eight π₯ minus sin nine π₯ to the power of eight.
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So, weβve seen within this question that we can apply the chain rule more than once within the same problem.
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In fact, we can apply it as many times as is necessary.
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Letβs remind ourselves then of some of the key points that weβve seen in this video.
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The chain rule is useful for differentiating composite functions, thatβs functions of other functions.
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If π¦ is equal to the composite function at π of π of π₯, then dπ¦ by dπ₯ is equal to π prime of π₯, thatβs the derivative of the inner function, multiplied by π prime of π of π₯.
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Thatβs the derivative of the outer function with the inner function still inside.
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Weβve also seen the that if we make the substitution π’ equals π of π₯, then π¦ becomes a function of π’.
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And the chain rule can be expressed as dπ¦ by dπ₯ is equal to dπ¦ by dπ’ multiplied by dπ’ by dπ₯.
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We find the derivative of π¦ with respect to π’ and multiply by the derivative of π’ with respect to π₯.
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We must make sure that we undo our substitution at the end, so that dπ¦ by dπ₯ is in terms of π₯ only.
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Weβve also seen the chain rule extension to the power rule, which tells us that the derivative of a function π of π₯ to the power of π is π prime of π₯ multiplied by π multiplied by π of π₯ to the π minus one.
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Finally, we saw that we can apply the chain rule as many times as we like within a particular problem.
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The chain rule is a really powerful tool.
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And it opens up a really wide class of functions which weβre able to differentiate.