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The Quotient Rule
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In this video, we will learn how to find the derivative of a function using the quotient rule.
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We will be looking at various examples of how it can be used.
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Consider the function π¦ is equal to negative three π₯ squared minus two π₯ plus 17 over the square root of π₯.
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If we wanted to find the derivative of this function, there are various methods which we could take.
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We could divide the numerator by the denominator and simply differentiate the resulting function.
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Alternatively, we could write the fraction as a product and find the derivative using the product rule.
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There is also an alternative method which we can use to find this derivative.
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And it requires no simplifying or rewriting of the equation.
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We call it the quotient rule.
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The derivation of the quotient rule is a bit long-winded for this video.
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So we wonβt be covering it here.
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The quotient rule says that given two differentiable functions, π’ of π₯ and π£ of π₯, the derivative of their quotient is given by π by dπ₯ of π’ of π₯ over π£ of π₯ is equal to π£ of π₯ times π by dπ₯ of π’ of π₯ minus π’ of π₯ times π by dπ₯ of π£ of π₯ all over π£ of π₯ squared.
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We can write this a lot more succinctly in prime notation.
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This gives us π’ over π£ prime is equal to π£π’ prime minus π’π£ prime all over π£ squared.
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I find an easy way to remember the quotient rule is with a rhyme.
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The rhyme goes LO π HI minus HI π LO over the square of whatβs below.
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Where HI is the numerator of the rational function which weβre differentiating.
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And LO is the denominator of this rational function.
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And the πs which are within the rhyme show where we need to differentiate.
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So π HI will be the differential of the numerator of our function.
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And π LO is the differential of the denominator of our function.
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You may find it easier to remember via another means.
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But feel free to use this method too.
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Weβre now ready to look at some examples.
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Find the first derivative of π¦ is equal to 8π₯ plus five over three π₯ plus 22.
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Here, we can see that our function π¦ is a rational function.
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So we can find its derivative by using the quotient rule.
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The quotient rule tells us that if we differentiate the quotient of two functions, so π’ over π£, with respect to π₯.
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Then itβs equal to π£ multiplied by the differential of π’ with respect to π₯ minus π’ timesed by the differential of π£ with respect to π₯ all over π£ squared.
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In order to find the first derivative of π¦, letβs start by labeling π’ and π£ from our equation.
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π’ will be equal to the numerator of the function, so eight π₯ plus five.
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And π£ will be equal to the denominator of the function, so thatβs three π₯ plus 22.
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Next, we must find d by dπ₯ of π’ and d by dπ₯ of π£, or dπ’ by dπ₯ and dπ£ by dπ₯.
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π’ and π£ are both polynomials.
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So we can simply differentiate them term by term.
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Writing π’ in terms of powers of π₯, we can say that itβs equal to eight π₯ to the power of one plus five π₯ to the power of zero.
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In order to differentiate, we simply multiply by the power and decrease the power by one.
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For the first time, we multiply by the power, so thatβs one, and decrease the power by one, to zero.
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Leaving us with one timesed by eight π₯ to the power of zero.
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For the second term, we multiply by the power, so thatβs zero, and decrease the power by one, to negative one.
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Giving us zero multiplied by five π₯ to the negative one.
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In the first term, π₯ to the power of zero is just one.
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So this becomes eight.
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In the second term, weβre multiplying by zero.
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So that term becomes zero.
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Therefore, we find that dπ’ by dπ₯ is equal to eight.
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We can use a similar method to find dπ£ by dπ₯.
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And we find that itβs equal to three.
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Now that we found dπ’ by dπ₯ and dπ£ by dπ₯, weβre ready to use the quotient rule.
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We find that dπ¦ by dπ₯ is equal to π£, which is three π₯ plus 22, multiplied by dπ’ by dπ₯, so thatβs eight, minus π’, so thatβs eight π₯ plus five, multiplied by dπ£ by dπ₯, so thatβs three.
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And this is all over π£ squared, so thatβs three π₯ plus 22 all squared.
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Next, we can expand the brackets.
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And then simplify to find that our solution is that the first derivative of π¦ is equal to 161 over three π₯ plus 22 all squared.
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Now, we will look at a slightly more complex example.
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Find the first derivative of the function π¦ is equal to four π₯ squared plus five π₯ plus five all over four π₯ squared minus two π₯ plus three.
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We can see that our function is a rational function.
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Therefore, we can use the quotient rule in order to find the derivative.
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The quotient rule tells us that π’ over π£ dash is equal to π£π’ dash minus π’π£ dash all over π£ squared.
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Where π’ is the numerator of our function and π£ is the denominator.
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In our case, π’ is equal to four π₯ squared plus five π₯ plus five.
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And π£ is equal to four π₯ squared minus two π₯ plus three.
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Now, we must find π’ prime and π£ prime.
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We do this by differentiating π’ and π£ with respect to π₯.
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Since π’ and π£ are both polynomial functions, we can find their derivatives by taking each term and multiplying the term by the power of π₯.
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And then, decreasing the power of π₯ by one.
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And doing this, we find that π’ prime is equal to eight π₯ plus five.
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And π£ prime is equal to eight π₯ minus two.
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Substituting these into our formula, we find that the first derivative of π¦ or π¦ prime is equal to π£ multiplied by π’ prime minus π’ multiplied by π£ prime all over π£ squared.
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Now, the result here looks quite daunting.
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However, we can still expand the brackets and then simplify.
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This is what we obtain after expanding the brackets in the numerator.
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Our final step is to simplify the numerator.
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Now, we have reached our solution.
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Which is that the first derivative of π¦ or π¦ prime is equal to negative 28π₯ squared minus 16π₯ plus 25 all over four π₯ squared minus two π₯ plus three all squared.
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Now, letβs look at a slightly different type of question.
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Suppose that π of π₯ is equal to π₯ squared plus ππ₯ plus π all over π₯ squared minus seven π₯ plus four.
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Given that π of zero is equal to one and π prime of zero is equal to four, find π and π.
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Our first step in this question can be to substitute π₯ equals zero into π of π₯.
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Since weβre given that π of zero is equal to one.
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We obtain that π of zero is equal to zero squared plus π times zero plus π all over zero squared minus seven times zero plus four.
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Now, all of these terms will go to zero apart from π and four.
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Weβre left with π of zero is equal to π over four.
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Next, we use the fact that the question has told us that π of zero is equal to one.
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And so, we can set this equal to one.
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From this, we find that π is equal to four.
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Next, we can use the fact that π prime of zero is equal to four.
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However, first of all, we must find π prime of π₯.
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In order to do this, we need to differentiate π.
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Since π is a rational function, we can use the quotient rule in order to find its derivative.
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The quotient rule tells us that π’ over π£ prime is equal to π£ times π’ prime minus π’ times π£ prime all over π£ squared.
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Setting our function π of π₯ equal to π’ over π£, we obtain that π’ is equal to π₯ squared plus ππ₯ plus π.
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And π£ is equal to π₯ squared minus seven π₯ plus four.
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We can find π’ prime and π£ prime by differentiating these two functions.
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Giving us that π’ prime is equal to two π₯ plus π and π£ prime is equal to two π₯ minus seven.
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Now, we can substitute these into the quotient rule.
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We obtain that π prime of π₯ is equal to π₯ squared minus seven π₯ plus four multiplied by two π₯ plus π minus π₯ squared plus ππ₯ plus π multiplied by two π₯ minus seven all over π₯ squared minus seven π₯ plus four all squared.
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Now, we could simplify π prime of π₯ at this point.
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However, weβre going to be substituting in π₯ is equal to zero.
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And so, a lot of these terms would just disappear.
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Letβs simply substitute π₯ equals zero in here.
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We obtain this.
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However, a lot of the terms will vanish to zero, which leaves us with four π plus seven π all over 16.
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Now, we have found that π is equal to four earlier.
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And so, we can substitute this in, giving us four π plus 28 all over 16.
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Since the question has told us that π prime of zero is equal to four, we can set this equal to four.
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Then, we simply rearrange this in order to solve for π.
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Now, we obtain our solution that π is equal to nine.
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Weβve now found the values of both π and π, which completes the solution to this question.
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In the next example, weβll be looking at a slightly different type of question.
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Let π of π₯ be equal to π of π₯ over negative four β of π₯ minus five.
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Given that π of negative two is equal to negative one, π prime of negative two is equal to negative eight, β of negative two is equal to negative two, and β prime of negative two is equal to five, find π prime of negative two.
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In this question, weβre asked to find π prime of negative two.
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So letβs start by differentiating π of π₯.
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π of π₯ is a rational function, so weβll need to use the quotient rule.
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The quotient rule tells us that π’ over π£ prime is equal to π£π’ prime minus π’π£ prime all over π£ squared.
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Setting π of π₯ equal to π’ over π£, we can see that π’ is equal to π of π₯.
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And π£ is equal to negative four β of π₯ minus five.
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π’ prime will be equal to π of π₯ prime.
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Now, the prime simply represents a differentiation with respect to π₯.
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So therefore, π of π₯ prime is identical to π prime of π₯.
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Next, we need to find π£ prime.
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So thatβs negative four β of π₯ minus five prime.
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Now, again, since a prime simply represents a differentiation with respect to π₯, we can apply normal differentiation rules here.
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And so, differentiating the constant term negative five will result in zero.
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So we can say that this is equal to negative four β of π₯ prime.
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Now, since our function β of π₯ is being multiplied by a constant, negative four.
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We can use our derivative rules and take the negative four out of the derivative.
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Giving us negative four multiplied by β of π₯ prime.
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And now, we can apply the same logic as we did for π of π₯ prime.
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And we can say that π£ prime is equal to negative four β prime of π₯.
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Now, we can substitute into the quotient rule in order to find π prime of π₯.
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Now that we have found π prime of π₯, we can substitute in π₯ is equal to negative two.
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Now, we have formed an equation in terms of π of negative two, π prime of negative two, β of negative two, and β prime of negative two.
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All of which we have been given the value of in the question.
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And so, weβre able to substitute in these values here.
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Now, our final step in finding π prime of negative two is to simplify this.
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Expanding the brackets, we get negative 24 minus 20 all over nine.
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This gives us a solution that π prime of negative two is equal to negative 44 over nine.
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Next, we will see how we can differentiate a function which consists of two rational expressions.
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If π¦ is equal to π₯ plus five over π₯ minus five minus π₯ minus five over π₯ plus five, find dπ¦ by dπ₯.
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Our function, π¦, consists of two rational expressions, π₯ plus five over π₯ minus five and π₯ minus five over π₯ plus five.
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And we could find dπ¦ by dπ₯ by using the quotient rule on these two rational expressions.
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However, this would require using the quotient rule twice.
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We can make our work a little easier by combining the two rational expressions into one.
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We obtain that π¦ is equal to π₯ plus five squared minus π₯ minus five squared all over π₯ minus five times π₯ plus five.
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We can expand the brackets and then simplify to obtain that π¦ is equal to 20π₯ over π₯ squared minus 25.
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And now, our function consists of only one rational expression.
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Weβre ready to use the quotient rule to differentiate this function.
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The quotient rule tells us that π’ over π£ prime is equal to π£π’ prime minus π’π£ prime over π£ squared.
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Setting π¦ equal to π’ over π£, we obtain that π’ is equal to 20π₯ and π£ is equal to π₯ squared minus 25.
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Next, we can find π’ prime and π£ prime, which gives us that π’ prime is equal to 20 and π£ prime is equal to two π₯.
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Now, we can substitute them into the quotient rule in order to find that dπ¦ by dπ₯ is equal to π₯ squared minus 25 times 20 minus 20π₯ times two π₯ all over π₯ squared minus 25 squared.
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We simplify this to obtain that dπ¦ by dπ₯ is equal to negative 20π₯ minus 500 all over π₯ squared minus 25 squared.
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In our final example, we will see how to evaluate the derivative of a rational function at a point.
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Evaluate π prime of three, where π of π₯ is equal to π₯ over π₯ plus two minus π₯ minus three over π₯ minus two.
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Now, our function is the difference of two rational expressions.
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We can start by combining the two rational expressions into one.
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We obtain that π of π₯ is equal to negative π₯ plus six all over π₯ squared minus four.
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And we have written π as a rational function.
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And weβre ready to use the quotient rule, which tells us that π’ over π£ prime is equal to π£π’ prime minus π’π£ prime all over π£ squared.
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Setting π of π₯ equal to π’ over π£, we obtain that π’ is equal to negative π₯ plus six, and π£ is equal to π₯ squared minus four.
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We then find that π’ prime is equal to negative one, and π£ prime is equal to two π₯.
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Substituting π’, π£, π’ prime, and π£ prime back into the quotient rule.
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We find that π prime of π₯ is equal to π₯ squared minus four multiplied by negative one minus negative π₯ plus six multiplied by two π₯ all over π₯ squared minus four squared.
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In order to find π prime of three, we simply substitute π₯ equals three into π prime of π₯.
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We obtain that π dash of three is equal to three squared minus four multiplied by negative one minus negative three plus six multiplied by two times three all over three squared minus four squared.
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Which simplifies to negative five minus 18 over 25.
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This gives us a solution that π prime of three is equal to negative 23 over 25.
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Now, we have seen a variety of examples of the quotient rule.
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Letβs cover some key points of the video.
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To find the derivative of the quotient of two differentiable functions, π’ of π₯ and π£ of π₯, we can use the quotient rule which states that d by dπ₯ of π’ of π₯ over π£ of π₯ is equal to π£ of π₯ times d by dπ₯ of π’ of π₯ minus π’ of π₯ times d by dπ₯ of π£ of π₯ all over π£ of π₯ squared.
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This is often written more succinctly using prime notation, as follows.
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π’ over π£ prime is equal to π£π’ prime minus π’π£ prime all over π£ squared.
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Before applying the quotient rule, it is worth checking whether itβs possible to simplify the expression for the function.
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This is particularly relevant when the function is expressed as the sum or difference of two rational expressions.