Video: Differentiating Rational Functions Using the Quotient Rule

Determine d𝑦/dπ‘₯, given that 𝑦 = ((π‘₯ βˆ’ 23)/(π‘₯ + 6))⁻⁡.

04:46

Video Transcript

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

Let’s have a look at this function whose derivative we’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. We’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, 𝑦, 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 𝑦 as π‘₯ plus six over π‘₯ minus 23 to the fifth power. We’ve changed the power from negative to positive and inverted the fraction. This isn’t absolutely necessary. But a lot of people prefer working with positive rather than negative exponents. Now, let’s think about answering the question. And what we can do first is define 𝑒 to be the entire function within the parentheses.

So we’re going to let 𝑒 equal π‘₯ plus six over π‘₯ minus 23. Then using our new definition of 𝑦, 𝑦 will be equal to 𝑒 to the fifth power. The reason this is helpful is because we now have 𝑦 as a function of 𝑒. And 𝑒 itself is a function of π‘₯. Which means we could apply the chain rule to find the derivative of 𝑦 with respect to π‘₯. The chain rule tells us that if 𝑦 is a function of 𝑒 and 𝑒 is a function of π‘₯, then d𝑦 by dπ‘₯ is equal to d𝑦 by d𝑒 multiplied by d𝑒 by dπ‘₯. We can find d𝑦 by d𝑒 using the power rule of differentiation. It’s equal to five 𝑒 to the fourth power. But what about d𝑒 by dπ‘₯?

Remember, 𝑒 is a quotient to functions, and they’re both differentiable functions. Which means we’re going to need to apply a second rule of differentiation, the quotient rule. This tells us that for two differentiable functions 𝑓 and 𝑔, the derivative of their quotient, 𝑓 over 𝑔, is equal to 𝑔 multiplied by 𝑓 prime minus 𝑓 multiplied by 𝑔 prime all over 𝑔 squared. To apply this rule, we therefore need to define 𝑓 to be the function in the numerator, π‘₯ plus six, and 𝑔 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, they’re both very straightforward. They’re each equal to one.

Substituting into the quotient rule then gives d𝑒 by dπ‘₯, or 𝑓 over 𝑔 prime, is equal to π‘₯ minus 23 multiplied by one, that’s 𝑔 multiplied by 𝑓 prime, minus π‘₯ plus six multiplied by one, that’s 𝑓 multiplied by 𝑔 prime, all over π‘₯ minus 23 squared, that’s 𝑔 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 d𝑒 by dπ‘₯ is equal to negative 29 over π‘₯ minus 23 squared.

So having now found both d𝑦 by d𝑒 and d𝑒 by dπ‘₯, we’re able to substitute into the chain rule to find d𝑦 by dπ‘₯. We have that d𝑦 by dπ‘₯ is equal to negative 29 over π‘₯ minus 23 squared multiplied by five 𝑒 to the fourth power. Now, this expression for d𝑦 by dπ‘₯ currently involves two variables, both π‘₯ and 𝑒. And we need it to be in terms of π‘₯ only. So the final step is to reverse the substitution we made earlier. 𝑒 was equal to π‘₯ plus six over π‘₯ minus 23. So replacing 𝑒 with this quotient, we now have that d𝑦 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 d𝑦 by dπ‘₯ for this function. It’s equal to negative 145 π‘₯ plus six to the fourth power over π‘₯ minus 23 to the sixth power.

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