### Video Transcript

If π¦ is equal to eight times the sec squared of five π₯ minus six, find dπ¦ by dπ₯.

We need to find an expression for the derivative of π¦ with respect to π₯ given that π¦ is equal to eight times the sec squared of five π₯ minus six. And thereβs several different ways we could do this. For example, we could rewrite eight times the sec squared of five π₯ as eight times the sec of five π₯ multiplied by the sec of five π₯. We can then evaluate this derivative by using the product rule. Similarly, since the sec squared of five π₯ is the composition of two functions, we take the sec of five π₯ and then square this entire expression. We could evaluate this by using the chain rule, and this would also work. However, weβre going to evaluate this by using the general power rule.

We recall the general power rule tells us for any 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₯ all raised to the power of π minus one. In fact, this is just an application of the chain rule. To use the general power rule, we need to notice the sec squared of five π₯ is the sec of five π₯ all squared.

And itβs worth pointing out this is the only derivative weβre going to need to calculate. This is because we know we can evaluate the derivative of π¦ with respect to π₯ term by term, and the derivative of the constant negative six is equal to zero. And, of course, we can take our constant factor of eight outside of our derivative. So to apply the general power rule on the sec of five π₯ all squared, weβll set π of π₯ to be our inner function, the sec of five π₯, and π to be our exponent of two. Of course, to apply the general power rule, we see we need to find an expression for π prime of π₯. Thatβs the derivative of the sec of five π₯ with respect to π₯.

And thereβs a few different ways we could evaluate this derivative. For example, we could rewrite the sec of five π₯ as one over the cos of five π₯ and then apply the general power rule or the quotient rule. However, itβs easier just to remember one of our standard trigonometric derivative results. For any constant π, the derivative of the sec of ππ₯ with respect to π₯ is equal to π times the tangent of ππ₯ multiplied by the sec of ππ₯. In our case, our value of π is equal to five. So we get that π prime of π₯ is equal to five times the tan of five π₯ multiplied by the sec of five π₯.

Weβre now ready to find an expression for the derivative of the sec squared of five π₯ with respect to π₯ by using the general power rule. Itβs equal to π times π prime of π₯ multiplied by π of π₯ all raised to the power of π minus one. Substituting in our values for π, π of π₯, and π prime of π₯, we get this derivative evaluates to give us two times five tan of five π₯ times the sec of five π₯ multiplied by the sec of five π₯ raised to the power of two minus one. And of course, we can simplify. First, in our exponent, two minus one simplifies to give us one, and raising something to the power of one doesnβt change its value.

Next, we have the sec of five π₯ multiplied by the sec of five π₯. So we can in fact simplify this to give us the sec squared of five π₯. And we can also simplify two minus five to give us 10. So this simplifies to give us 10 times the tan of five π₯ multiplied by the sec squared of five π₯. But weβre not yet done. Remember, weβve only differentiated the sec squared of five π₯. We need to differentiate our expression for π¦, so we need to find an expression for dπ¦ by dπ₯. Thatβs the derivative of eight times the sec squared of five π₯ minus six with respect to π₯.

First, weβd evaluate this derivative term by term. Thatβs the derivative of eight times the sec squared of five π₯ with respect to π₯ minus the derivative of six with respect to π₯. Of course, six is a constant, so its derivative with respect to π₯ is equal to zero. Next, weβll use our laws of derivatives to take the constant factor of eight outside of our derivative. And we already found an expression for the derivative of the sec squared of five π₯ with respect to π₯. Itβs equal to 10 tan of five π₯ times the sec squared of five π₯.

So if we substitute this into our expression, we get eight times 10 tan five π₯ multiplied by the sec squared of five π₯. And finally, we can simplify eight times 10 to give us 80. Therefore, by using the general power rule, we were able to show if π¦ is equal to eight times the sec squared of five π₯ minus six, then the derivative of π¦ with respect to π₯ is equal to 80 times the tan of five π₯ multiplied by the sec squared of five π₯.