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In this video, weβre going to see how three rules of differentiation can be combined to allow us to differentiate increasingly complex functions.
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We remind ourselves of these three rules, the product rule, the quotient rule, and the chain rule.
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And then, weβll apply combinations of these rules to a couple of different examples.
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First, letβs recall these three rules and their uses.
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The first rule is the product rule, which allows us to differentiate products of functions.
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It tells us that the derivative of the product ππ, thatβs ππ prime, is equal to ππ prime plus π prime π.
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What weβre doing is multiplying each function by the derivative of the other and adding them together.
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The second rule is the quotient rule, which allows us to differentiate quotients for functions, thatβs π over π.
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And it tells us that the derivative of π over π is equal to ππ prime minus ππ prime over π squared.
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Now, itβs important to note that unlike the product rule, which is symmetrical in π and π, the quotient rule is not symmetrical in π and π.
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So, we must make sure that we define π to be the function in the numerator and π to be the function in the denominator.
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The third rule is the chain rule, which allows us to differentiate composite functions; that is, functions of other functions.
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Here, we have the function π of π of π₯, which means we apply π first and then apply π.
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Its derivative is given by π prime of π₯ multiplied by π prime of π of π₯.
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Thatβs the derivative of the inner function multiplied by the derivative of the outer function, with the inner function still inside it.
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It is important that weβre clear on the different uses of notation.
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In the product rule, ππ means π multiplied by π.
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Whereas in the chain rule, π of π of π₯ means the composite function we get when we apply π first and then apply π.
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We can also express each of these rules using Leibnizβs notation.
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And itβs more common to use the letters π’ and π£ when we do so, although it doesnβt actually matter.
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The product rule tells us that the derivative with respect to π₯ of π’π£ is equal to π’ dπ£ by dπ₯ plus π£ dπ’ by dπ₯.
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The quotient rule tells us that the derivative with respect to π₯ of π’ over π£ is equal to π£ dπ’ by dπ₯ minus π’ dπ£ by dπ₯ all over π£ squared.
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And finally, the chain rule tells us that if π¦ is equal to the composite function π of π of π₯, and we define π’ to be π of π₯ so that π¦ becomes a function of π’, π¦ equals π of π’.
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Then, dπ¦ by dπ₯ is equal to dπ¦ by dπ’ multiplied by dπ’ by dπ₯.
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We differentiate π¦ with respect to π’ and then multiply by the derivative of π’ with respect to π₯.
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Weβll now apply combinations of these three rules to some examples.
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Find the first derivative of π¦ equals π₯ minus five multiplied by π₯ minus two to the power of six at one, negative four.
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Letβs begin then by considering this function that weβve been given.
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We can see that it is a product of two functions, π₯ minus five and π₯ minus two to the power of six.
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This suggests that weβre going to need to use the product rule in order to find this derivative.
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So, we can define one function to be π and the other function to be π.
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Here is the product rule but as itβs symmetrical in π and π, it doesnβt actually matter which way round we make this definition.
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We have then that π of π₯ is equal to π₯ minus five and π of π₯ is equal to π₯ minus two to the power of six.
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And in order to apply the product rule, we need to find the derivatives of π and π.
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π prime of π₯ is straightforward.
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Itβs just one.
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But what about π prime?
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We have a power of six, and we certainly donβt want to distribute all of the parentheses to give a polynomial.
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So, how are we going to find this derivative?
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π is actually a composite function, which means that we can apply the chain rule in order to find its derivative, dπ¦ by dπ₯ equals dπ¦ by dπ’ multiplied by dπ’ by dπ₯.
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We can amend the π¦s in the chain rule to πs.
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And then, we can allow π’ to equal π₯ minus two, which makes π equal to π’ to the power of six.
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The derivative of π’ with respect to π₯ is one.
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And applying the power rule, the derivative of π with respect to π’ is six π’ to the power of five.
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Applying the chain rule, dπ by dπ₯ is equal to six π’ to the power of five, thatβs dπ by dπ’, multiplied by one, thatβs dπ’ by dπ₯.
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But of course, multiplying by one has no effect.
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So, dπ by dπ₯ is equal to six π’ to the power of five.
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However, we need this derivative to be in terms of π₯, so we must reverse the substitution.
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π’ is equal to π₯ minus two, so dπ by dπ₯ in terms of π₯ is equal to six multiplied by π₯ minus two to the power of five.
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We could also have seen this by applying the chain rule extension to the power rule.
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Now that weβve found each derivative, we can begin substituting into the product rule to find the derivative of π¦.
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dπ¦ by dπ₯ will be equal to π, thatβs π₯ minus five, multiplied by π prime, thatβs six multiplied by π₯ minus two to the power of five.
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We then add π prime, thatβs just one, multiplied by π, which is π₯ minus two to the power of six.
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We can simplify by taking a common factor of π₯ minus two to the power of five out.
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And then, simplifying in that second bracket gives π₯ minus two to the power of five multiplied by seven π₯ minus 32.
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So, weβve found the first derivative of π¦.
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But we are asked to evaluate this derivative at a particular point, the point one, negative four.
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This means that we need to make a substitution.
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We need to substitute the π₯-value at this point into our gradient function, giving one minus two to the power of five multiplied by seven multiplied by one minus 32.
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That gives negative one to the power of five multiplied by negative 25, which is 25.
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So in this question, we required a combination of the product and chain rules in order to answer this problem.
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Next, weβll consider an example which requires a different combination of rules.
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Determine the derivative of the function π of π’ equals π’ squared plus five over π’ squared minus one all to the power of four.
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Letβs begin by considering the rules that will be useful to us in this question.
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We can see that we have a quotient, π’ squared plus five over π’ squared minus one.
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So, weβre going to need to use the quotient rule.
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Iβve written the quotient rule out here using πs and πs because weβve already got π’s and πs in the question.
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So, this will enable us to find the derivative of this expression inside the parentheses.
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But we still have that power of four.
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Weβll allow π£ to equal the expression inside the parentheses.
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Itβs equal to π’ squared plus five over π’ squared minus one.
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Then, π will be equal to π£ to the power of four.
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And we can apply the chain rule, dπ by dπ’ is equal to dπ by dπ£ multiplied by dπ£ by dπ’.
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dπ by dπ£ can be easily calculated by applying the power rule.
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Itβs equal to four π£ cubed.
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But in order to calculate dπ£ by dπ’, weβre going to need to apply the quotient rule.
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Weβll let π equal the numerator of π£, thatβs π’ squared plus five, and π equals the denominator, thatβs π’ squared minus one.
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dπ by dπ’ and dπ by dπ’ can be found by applying the power rule.
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Theyβre each equal to two π’.
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Now, we can substitute into the quotient rule to find dπ£ by dπ’.
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We have π, thatβs π’ squared minus one, multiplied by π prime or dπ by dπ’, thatβs two π’, minus π, thatβs π’ squared plus five, multiplied by π prime or dπ by dπ’, thatβs two π’.
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Itβs all divided by π squared, thatβs π’ squared minus one squared.
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Now, letβs simplify.
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Expanding the parentheses in the numerator, we have two π’ cubed minus two π’ minus two π’ cubed minus 10π’.
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And then the denominator remains π’ squared minus one all squared.
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The two π’ cubeds will cancel each other out, leaving negative 12π’ over π’ squared minus one squared.
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Weβll now delete some of this working out for the quotient rule to make some room on the page.
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We now have then that dπ by dπ£ is equal to four π£ cubed and dπ£ by dπ’ is equal to negative 12π’ over π’ squared minus one all squared.
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So, we can substitute into the chain rule.
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dπ by dπ’ is equal to four π£ cubed multiplied by negative 12π’ over π’ squared minus one all squared.
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Now, remember that dπ by dπ’ must be in terms of π’ only.
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So, we need to reverse our substitution.
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π£ is equal to π’ squared plus five over π’ squared minus one.
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So, we have four π’ squared plus five over π’ squared minus one all cubed multiplied by negative 12π’ over π’ squared minus one all squared.
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Simplifying gives negative 48π’ multiplied by π’ squared plus five cubed all over π’ squared minus one to the power of five.
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In this question, we saw then that we needed to apply a combination of the quotient and chain rules in order to find the derivative of the function π of π’.
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Letβs now look at an example involving trigonometric functions.
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Determine the derivative of the function π of π‘ equals the square root of negative sin π‘ plus seven over negative cos π‘ plus seven.
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Now, we can see straightaway in this question that we have a quotient.
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So, weβre going to need to apply the quotient rule at some point.
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But are we going to need to do anything else?
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Well, we donβt just have this quotient.
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We have the square root of this quotient, which means we have a composite function.
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And so, weβre also going to need to apply the chain rule.
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Weβll begin by allowing π’ to be that quotient underneath the square root.
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π’ is equal to negative sin π‘ plus seven over negative cos π‘ plus seven.
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Then, π becomes a function of π’.
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Itβs 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, using the letters π , π‘, and π’ as we have in this question, tells us that the derivative of π with the respect to π‘ is equal to dπ by dπ’ multiplied by dπ’ by dπ‘.
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Applying the power rule, we see that dπ by dπ’ is equal to one-half π’ to the power of negative a half.
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But in order to find dπ’ by dπ‘, weβre going to need to apply the quotient rule.
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Weβll define π to be the function in the numerator, thatβs negative sin π‘ plus seven, and π to be the function in the denominator, negative cos π‘ plus seven.
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In order to find the derivatives of these two functions, we need to recall how we differentiate sin and cos.
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There is a helpful little cycle that we can remember.
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The derivative of sin π‘ is cos π‘.
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The derivative of cos π‘ is negative sin π‘.
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The derivative of negative sin π‘ is negative cos π‘.
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And the derivative of negative cos π‘ is sin π‘.
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And then we go around the cycle again.
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Remembering that the derivative of a constant is just zero, we have that π prime is equal to negative cos π‘ and π prime is equal to sin π‘.
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Now, we can substitute into the quotient rule to find dπ’ by dπ‘.
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Itβs π, thatβs negative cos π‘ plus seven, multiplied by π prime, negative cos π‘, minus π, thatβs negative sin π‘ plus seven, multiplied by π prime, thatβs sin π‘, all over π squared.
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Now, we need to do some simplification.
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So, weβll expand the parentheses in the numerator, which gives cos squared π‘ minus seven cos π‘ plus sin squared π‘ minus seven sin π‘ over negative cos π‘ plus seven squared.
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We can recall, at this point, one of our trigonometric identities, cos squared π‘ plus sin squared π‘ is equal to one.
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So, this simplifies to one minus seven cos π‘ minus seven sin π‘ over negative cos π‘ plus seven squared.
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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 dπ by dπ’, thatβs a half π’ to the power of negative a half, multiplied by dπ’ by dπ‘, thatβs one minus seven cos π‘ minus seven sin π‘ over negative cos π‘ plus seven squared.
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Remember, though, that dπ by dπ‘ must be in terms of π‘ only.
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So, we need to reverse our substitution.
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We have negative sin π‘ plus seven over negative cos π‘ plus seven to the power of negative a half multiplied by one minus seven cos π‘ minus seven sin π‘ over two multiplied by negative cos π‘ plus seven squared.
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That power of negative a half means a reciprocal, so we can deal with that by inverting the fraction.
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And the first part becomes negative cos π‘ plus seven over negative sin π‘ plus seven to the power of positive one-half.
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We can then simplify the powers.
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We have negative cos π‘ plus seven to the power of a half in the numerator and then negative cos π‘ plus seven to the power of two in the denominator.
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Which will lead to a power of negative three over two overall.
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Thatβs a power of three over two in the denominator.
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This leads us to one minus seven cos π‘ minus seven sin π‘ in the numerator.
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And in the denominator two times the square root of negative sin π‘ plus seven.
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Thatβs negative sin π‘ plus seven to the power of a half multiplied by negative cos π‘ plus seven to the power of three over two.
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In this question then, weβve seen that we can apply both the quotient and chain rule to a problem involving the derivatives of trigonometric functions.
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Letβs now summarize what weβve seen in this video.
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Weβve reminded ourselves of these three key rules, the product rule, the quotient rule, and the chain rule, each expressed here using Leibnizβs notation.
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Weβve seen that these three rules can be used together in order to find the derivatives of more complex functions.
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And although we havenβt done an example in this video, we can, of course, apply each rule multiple times if the problem requires it.
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These three rules are incredibly powerful.
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And by combining them or using the same rule in succession, this opens up a wide class of complex functions whose derivatives weβre now able to find.