### Video Transcript

Find dπ¦ by dπ₯ given that π¦ is equal to negative six π₯ squared plus four all divided by two π₯ squared minus five all raised to the seventh power.

We need to find an expression for dπ¦ by dπ₯ given the following expression for π¦. And thereβs a lot of different ways of doing this. Weβre only going to go through one of these. Weβll call our inner rational function π of π₯. So by taking π of π₯ to be our inner rational function, we can see that π¦ is equal to π of π₯ all raised to the seventh power.

So we can find an expression for dπ¦ by dπ₯ by either using the chain rule or the general power rule. Weβll do this by using the general power rule. We recall this tells us for any real constant π and differentiable function π of π₯, the derivative of π of π₯ all raised to the πth power with respect to π₯ is equal to π times π prime of π₯ multiplied by π of π₯ raised to the power of π minus one.

In our case, our exponent π is seven and our inner function is π of π₯. And itβs worth pointing out we know π of π₯ is differentiable because itβs a rational function. We could evaluate the derivative by using the quotient rule. So by setting π equal to seven and π of π₯ to be our inner function π of π₯, we get dπ¦ by dπ₯ is equal to seven times π prime of π₯ multiplied by π of π₯ all raised to the sixth Power. And we have an expression for π of π₯. Itβs our inner rational function. So we can substitute our expression for π of π₯ into this expression for dπ¦ by dπ₯.

Now, we can see the only thing we need to find dπ¦ by dπ₯ is to find an expression for π prime of π₯. And thatβs the derivative of a rational function. We know we can do this by using the quotient rule. So to find an expression for π prime of π₯, we need to start by recalling the quotient rule. We recall if π’ and π£ are differentiable functions, the derivative of π’ over π£ is equal to π’ prime π£ minus π£ prime π’ all divided by π£ squared.

Remember, weβre using the quotient rule to find the derivative of our inner rational function π of π₯. So we set π’ of π₯ to be its numerator, negative six π₯ squared plus four, and π£ of π₯ to be its denominator, two π₯ squared minus five. And of course, to use the quotient rule, we need to find expressions for π’ prime of π₯ and π£ prime of π₯. In fact, we can evaluate both of these derivatives by using the power rule for differentiation. We get π’ prime of π₯ is negative 12π₯ and π£ prime of π₯ is equal to four π₯.

Weβre now ready to find an expression for π prime of π₯ by using the quotient rule. Itβs the derivative of π’ of π₯ over π£ of π₯ with respect to π₯. Substituting in our expressions for π’, π£, π’ prime, and π£ prime into our quotient rule formula, we get π prime of π₯ is equal to negative 12π₯ times two π₯ squared minus five minus four π₯ multiplied by negative six π₯ squared plus four all divided by two π₯ squared minus five all squared.

And we can simplify this expression. Weβll start by distributing negative 12π₯ over our first set of parentheses. Doing this gives us negative 24π₯ cubed plus 60π₯. Next, weβll distribute negative four π₯ over our second set of parentheses. This gives us 24π₯ cubed minus 16π₯ squared. And then, of course, we still need to divide through by two π₯ squared minus five all squared. And we can see in our numerator we can cancel negative 24π₯ cubed plus 24π₯ cubed. And we can simplify 60π₯ minus 16π₯ is equal to 44π₯.

So we were able to simplify this expression to give us 44π₯ divided by two π₯ squared minus five all squared. And remember, this is an expression for π prime of π₯. So we can substitute this into the expression we got for dπ¦ by dπ₯. Substituting our expression for π prime of π₯ into our expression for dπ¦ by dπ₯ gives us the following expression. And we could leave our answer like this. However, we can also simplify slightly.

First, weβll multiply 44π₯ by seven. Doing this gives us 308. Next, in our second factor, instead of raising the entire fraction to the sixth power, weβre going to raise the numerator and the denominator separately to the sixth power. And by doing this, we get the following expression.

Now, we can multiply these two together. In our denominator, we have two π₯ squared minus five all squared multiplied by two π₯ squared minus five all raised to the sixth power. By using our laws of exponents, this is two π₯ squared minus five all raised to the eighth power. And by doing this, we get our final answer.

Therefore, we were able to show if π¦ is equal to negative six π₯ squared plus seven all divided by two π₯ squared minus five all raised to the seventh power, then by using the general power rule and the quotient rule, we were able to show dπ¦ by dπ₯ is equal to 308π₯ times negative six π₯ squared plus four raised to the sixth power all over two π₯ squared minus five all raised to the eighth power.