Determine d𝑦 by d𝑥, given that 𝑦 equals 𝑥 minus 23 over 𝑥 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, 𝑥 minus 23 over 𝑥 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 𝑥 plus six over 𝑥 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 𝑢 to be the entire function within the parentheses.
So we’re going to let 𝑢 equal 𝑥 plus six over 𝑥 minus 23. Then using our new definition of 𝑦, 𝑦 will be equal to 𝑢 to the fifth power. The reason this is helpful is because we now have 𝑦 as a function of 𝑢. And 𝑢 itself is a function of 𝑥. Which means we could apply the chain rule to find the derivative of 𝑦 with respect to 𝑥. The chain rule tells us that if 𝑦 is a function of 𝑢 and 𝑢 is a function of 𝑥, then d𝑦 by d𝑥 is equal to d𝑦 by d𝑢 multiplied by d𝑢 by d𝑥. We can find d𝑦 by d𝑢 using the power rule of differentiation. It’s equal to five 𝑢 to the fourth power. But what about d𝑢 by d𝑥?
Remember, 𝑢 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, 𝑥 plus six, and 𝑔 to be the function in the denominator, 𝑥 minus 23. We can find each of their derivatives with respect to 𝑥 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 d𝑢 by d𝑥, or 𝑓 over 𝑔 prime, is equal to 𝑥 minus 23 multiplied by one, that’s 𝑔 multiplied by 𝑓 prime, minus 𝑥 plus six multiplied by one, that’s 𝑓 multiplied by 𝑔 prime, all over 𝑥 minus 23 squared, that’s 𝑔 squared. Distributing the parentheses in the numerator gives 𝑥 minus 23 minus 𝑥 minus six over 𝑥 minus 23 all squared. And finally, collecting like terms gives that d𝑢 by d𝑥 is equal to negative 29 over 𝑥 minus 23 squared.
So having now found both d𝑦 by d𝑢 and d𝑢 by d𝑥, we’re able to substitute into the chain rule to find d𝑦 by d𝑥. We have that d𝑦 by d𝑥 is equal to negative 29 over 𝑥 minus 23 squared multiplied by five 𝑢 to the fourth power. Now, this expression for d𝑦 by d𝑥 currently involves two variables, both 𝑥 and 𝑢. And we need it to be in terms of 𝑥 only. So the final step is to reverse the substitution we made earlier. 𝑢 was equal to 𝑥 plus six over 𝑥 minus 23. So replacing 𝑢 with this quotient, we now have that d𝑦 by d𝑥 is equal to negative 29 over 𝑥 minus 23 squared multiplied by five 𝑥 plus six over 𝑥 minus 23 to the power of four.
In the numerator, negative 29 multiplied by five gives negative 145. And then, we have 𝑥 plus six to the fourth power. And in the denominator, we have 𝑥 minus 23 squared multiplied by 𝑥 minus 23 to the fourth power. So adding the exponents gives 𝑥 minus 23 to the sixth power. So by applying both the chain and quotient rules, we found d𝑦 by d𝑥 for this function. It’s equal to negative 145 𝑥 plus six to the fourth power over 𝑥 minus 23 to the sixth power.