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

Given that π¦ = β6π§Β² β 23 and π§ = β(4/π₯), determine ππ¦/ππ₯.

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### Video Transcript

Given that π¦ equals negative six π§ squared minus 23 and π§ is equal to negative four of π₯, determine ππ¦ by ππ₯.

Weβre looking for the derivative of π¦ with respect to π₯ and weβre given π¦ in terms of another variable π§ and π§ in terms of π₯. So this looks like a job for the chain rule, which says that the derivative of π¦ with respect to π₯ is equal to the derivative of π¦ with respect to some other variable π§ times the derivative of π§ with respect to π₯.

Letβs apply the chain rule to our problem. We need to find ππ¦ by ππ§. That is the derivative of π¦ with respect to π§. Luckily, we have π¦ written in terms of π§. So this is straightforward. π¦ equals negative six π§ squared minus 23. We differentiate both sides with respect to π§. 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 12π§.

Having found ππ¦ by ππ§, we now move on to finding ππ§ by ππ₯. And to find ππ§ by ππ₯, we use the relation between π§ and π₯. π§ equals negative four over π₯ and we can write this in exponent notation as π§ 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, weβll change the π§ in our rule to π₯. Itβs important to note that we could replace π§ by any letter that we would like and it would still express the same rule. But we choose to replace π§ by π₯ because weβre going to be differentiating with respect to π₯. Applying our rule, we find ππ§ by ππ₯ to be four π₯ to the negative two or if you want to avoid negative exponents four over π₯ squared.

Using this expression for ππ§ by ππ₯, we see that ππ¦ by ππ₯ is negative 12 times π§ times four over π₯ squared. Now, itβs just a case of simplifying this expression. And as part of the simplification, we realize that we have ππ¦ by ππ₯ written not only in terms of π₯, but also in terms of the variable π§. And weβd like it to be solely in terms of π₯ if thatβs possible. And it is possible because π§ is negative four over π₯.

Substituting negative four over π₯ for π§, we get an expression for ππ¦ by ππ₯ 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.

ππ¦ by ππ₯ is then 192 over π₯ cubed.