Video Transcript
Determine dπ¦ by dπ₯, given that π¦ equals π₯ minus 23 over π₯ plus six to the power of negative five.
Letβs have a look at this function whose derivative weβve 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. Weβre 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. Weβve changed the power from negative to positive and inverted the fraction. This isnβt absolutely necessary. But a lot of people prefer working with positive rather than negative exponents. Now, letβs think about answering the question. And what we can do first is define π’ to be the entire function within the parentheses.
So weβre 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. Itβs equal to five π’ to the fourth power. But what about dπ’ by dπ₯?
Remember, π’ is a quotient to functions, and theyβre both differentiable functions. Which means weβre 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, theyβre both very straightforward. Theyβre 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, thatβs π multiplied by π prime, minus π₯ plus six multiplied by one, thatβs π multiplied by π prime, all over π₯ minus 23 squared, thatβs 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₯, weβre 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. Itβs equal to negative 145 π₯ plus six to the fourth power over π₯ minus 23 to the sixth power.
