Question Video: Differentiating Functions Involving Trigonometric Ratios Using the Chain Rule

If 𝑦 = 8 secΒ² 5π‘₯ βˆ’ 6, find d𝑦/dπ‘₯.

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Video Transcript

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

We need to find an expression for the derivative of 𝑦 with respect to π‘₯ given that 𝑦 is equal to eight times the sec squared of five π‘₯ minus six. And there’s 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, we’re going to evaluate this by using the general power rule.

We recall the general power rule tells us for any constant 𝑛 and differentiable function 𝑔 of π‘₯, the derivative of 𝑔 of π‘₯ all raised to the 𝑛th power with respect to π‘₯ is equal to 𝑛 times 𝑔 prime of π‘₯ multiplied by 𝑔 of π‘₯ all raised to the power of 𝑛 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 it’s worth pointing out this is the only derivative we’re going to need to calculate. This is because we know we can evaluate the derivative of 𝑦 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, we’ll set 𝑔 of π‘₯ to be our inner function, the sec of five π‘₯, and 𝑛 to be our exponent of two. Of course, to apply the general power rule, we see we need to find an expression for 𝑔 prime of π‘₯. That’s the derivative of the sec of five π‘₯ with respect to π‘₯.

And there’s 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, it’s easier just to remember one of our standard trigonometric derivative results. For any constant π‘Ž, the derivative of the sec of π‘Žπ‘₯ with respect to π‘₯ is equal to π‘Ž times the tangent of π‘Žπ‘₯ multiplied by the sec of π‘Žπ‘₯. In our case, our value of π‘Ž is equal to five. So we get that 𝑔 prime of π‘₯ is equal to five times the tan of five π‘₯ multiplied by the sec of five π‘₯.

We’re 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. It’s equal to 𝑛 times 𝑔 prime of π‘₯ multiplied by 𝑔 of π‘₯ all raised to the power of 𝑛 minus one. Substituting in our values for 𝑛, 𝑔 of π‘₯, and 𝑔 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 doesn’t 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 we’re not yet done. Remember, we’ve only differentiated the sec squared of five π‘₯. We need to differentiate our expression for 𝑦, so we need to find an expression for d𝑦 by dπ‘₯. That’s the derivative of eight times the sec squared of five π‘₯ minus six with respect to π‘₯.

First, we’d evaluate this derivative term by term. That’s 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, we’ll 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‘₯. It’s 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 𝑦 is equal to eight times the sec squared of five π‘₯ minus six, then the derivative of 𝑦 with respect to π‘₯ is equal to 80 times the tan of five π‘₯ multiplied by the sec squared of five π‘₯.

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