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Question Video: Finding the First Derivative of a Composite Function Using the Chain Rule at a Point Mathematics • Higher Education

Given that 𝑦 = βˆ’6𝑧² βˆ’ 23 and 𝑧 = βˆ’(4/π‘₯), determine 𝑑𝑦/𝑑π‘₯.


Video Transcript

Given that 𝑦 equals negative six 𝑧 squared minus 23 and 𝑧 is equal to negative four of π‘₯, determine 𝑑𝑦 by 𝑑π‘₯.

We’re looking for the derivative of 𝑦 with respect to π‘₯ and we’re given 𝑦 in terms of another variable 𝑧 and 𝑧 in terms of π‘₯. So this looks like a job for the chain rule, which says that the derivative of 𝑦 with respect to π‘₯ is equal to the derivative of 𝑦 with respect to some other variable 𝑧 times the derivative of 𝑧 with respect to π‘₯.

Let’s apply the chain rule to our problem. We need to find 𝑑𝑦 by 𝑑𝑧. That is the derivative of 𝑦 with respect to 𝑧. Luckily, we have 𝑦 written in terms of 𝑧. So this is straightforward. 𝑦 equals negative six 𝑧 squared minus 23. We differentiate both sides with respect to 𝑧. Using the fact that the derivative of a difference of functions is the difference of their derivatives and using the formula for the derivative of a number times a power with respect to the base of that power and also the fact that the derivative of a constant function is zero, we get negative 12𝑧.

Having found 𝑑𝑦 by 𝑑𝑧, we now move on to finding 𝑑𝑧 by 𝑑π‘₯. And to find 𝑑𝑧 by 𝑑π‘₯, we use the relation between 𝑧 and π‘₯. 𝑧 equals negative four over π‘₯ and we can write this in exponent notation as 𝑧 equals negative four π‘₯ to the negative one. We can now apply our rule about the derivative of a number times a power of a variable with respect to that variable.

To make things clearer, we’ll change the 𝑧 in our rule to π‘₯. It’s important to note that we could replace 𝑧 by any letter that we would like and it would still express the same rule. But we choose to replace 𝑧 by π‘₯ because we’re going to be differentiating with respect to π‘₯. Applying our rule, we find 𝑑𝑧 by 𝑑π‘₯ to be four π‘₯ to the negative two or if you want to avoid negative exponents four over π‘₯ squared.

Using this expression for 𝑑𝑧 by 𝑑π‘₯, we see that 𝑑𝑦 by 𝑑π‘₯ is negative 12 times 𝑧 times four over π‘₯ squared. Now, it’s just a case of simplifying this expression. And as part of the simplification, we realize that we have 𝑑𝑦 by 𝑑π‘₯ written not only in terms of π‘₯, but also in terms of the variable 𝑧. And we’d like it to be solely in terms of π‘₯ if that’s possible. And it is possible because 𝑧 is negative four over π‘₯.

Substituting negative four over π‘₯ for 𝑧, we get an expression for 𝑑𝑦 by 𝑑π‘₯ in terms of π‘₯ alone. And we can simplify this expression with negative 12 times negative four times four making 192 in the numerator and π‘₯ times π‘₯ squared making π‘₯ cubed in the denominator.

𝑑𝑦 by 𝑑π‘₯ is then 192 over π‘₯ cubed.

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