# Question Video: Differentiating Rational Functions Using the Quotient Rule Mathematics • Higher Education

Determine dπ¦/dπ₯, given that π¦ = ((π₯ β 23)/(π₯ + 6))β»β΅.

04:46

### Video Transcript

Determine dπ¦ by dπ₯, given that π¦ equals π₯ minus 23 over π₯ plus six to the power of negative five.

Letβs have a look at this function whose derivative weβve been asked to find. We can see that it includes a quotient, π₯ minus 23 over π₯ plus six. And then the entire quotient is raised to the power of negative five. So we have a composite function, a function of a function. Weβre going to need to use a couple of different rules of differentiation in order to answer this problem. First, though, we can simplify our function, π¦, a little if we wish.

We know that a negative exponent defines a reciprocal. And in the case of a fraction, to find its reciprocal, we just invert that fraction. We swap the numerator and denominator round. So we can write π¦ as π₯ plus six over π₯ minus 23 to the fifth power. Weβve changed the power from negative to positive and inverted the fraction. This isnβt absolutely necessary. But a lot of people prefer working with positive rather than negative exponents. Now, letβs think about answering the question. And what we can do first is define π’ to be the entire function within the parentheses.

So weβre going to let π’ equal π₯ plus six over π₯ minus 23. Then using our new definition of π¦, π¦ will be equal to π’ to the fifth power. The reason this is helpful is because we now have π¦ as a function of π’. And π’ itself is a function of π₯. Which means we could apply the chain rule to find the derivative of π¦ with respect to π₯. 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₯. We can find dπ¦ by dπ’ using the power rule of differentiation. Itβs equal to five π’ to the fourth power. But what about dπ’ by dπ₯?

Remember, π’ is a quotient to functions, and theyβre both differentiable functions. Which means weβre going to need to apply a second rule of differentiation, the quotient rule. This tells us that for two differentiable functions π and π, the derivative of their quotient, π over π, is equal to π multiplied by π prime minus π multiplied by π prime all over π squared. To apply this rule, we therefore need to define π to be the function in the numerator, π₯ plus six, and π to be the function in the denominator, π₯ minus 23. We can find each of their derivatives with respect to π₯ using the power rule of differentiation. And in fact, theyβre both very straightforward. Theyβre each equal to one.

Substituting into the quotient rule then gives dπ’ by dπ₯, or π over π prime, is equal to π₯ minus 23 multiplied by one, thatβs π multiplied by π prime, minus π₯ plus six multiplied by one, thatβs π multiplied by π prime, all over π₯ minus 23 squared, thatβs π squared. Distributing the parentheses in the numerator gives π₯ minus 23 minus π₯ minus six over π₯ minus 23 all squared. And finally, collecting like terms gives that dπ’ by dπ₯ is equal to negative 29 over π₯ minus 23 squared.

So having now found both dπ¦ by dπ’ and dπ’ by dπ₯, weβre able to substitute into the chain rule to find dπ¦ by dπ₯. We have that dπ¦ by dπ₯ is equal to negative 29 over π₯ minus 23 squared multiplied by five π’ to the fourth power. Now, this expression for dπ¦ by dπ₯ currently involves two variables, both π₯ and π’. And we need it to be in terms of π₯ only. So the final step is to reverse the substitution we made earlier. π’ was equal to π₯ plus six over π₯ minus 23. So replacing π’ with this quotient, we now have that dπ¦ by dπ₯ is equal to negative 29 over π₯ minus 23 squared multiplied by five π₯ plus six over π₯ minus 23 to the power of four.

In the numerator, negative 29 multiplied by five gives negative 145. And then, we have π₯ plus six to the fourth power. And in the denominator, we have π₯ minus 23 squared multiplied by π₯ minus 23 to the fourth power. So adding the exponents gives π₯ minus 23 to the sixth power. So by applying both the chain and quotient rules, we found dπ¦ by dπ₯ for this function. Itβs equal to negative 145 π₯ plus six to the fourth power over π₯ minus 23 to the sixth power.