Video: Differentiating a Combination of Trigonometric Functions at a Point

If 𝑦 = 8 cot π‘₯ + 5 sec π‘₯ , find 𝑑𝑦/𝑑π‘₯ at π‘₯ = πœ‹/6.

08:06

Video Transcript

If 𝑦 equals eight cot π‘₯ plus five sec π‘₯, find 𝑑𝑦 by 𝑑π‘₯ at π‘₯ equals πœ‹ by six.

We’re looking for the derivative of 𝑑𝑦 by 𝑑π‘₯. Well, actually we’re looking for the value of 𝑑𝑦 by 𝑑π‘₯ at π‘₯ equals πœ‹ by six. But to find this, we first need to find 𝑑𝑦 by 𝑑π‘₯ in terms of π‘₯. So let’s differentiate. We’re looking for the derivative of eight cot π‘₯ plus five sec π‘₯. And we use the fact that the derivative of a sum of functions is the sum of their derivatives and the fact that the derivative of a number times a function is that number times the derivative of the function to write this derivative 𝑑𝑦 by 𝑑π‘₯ in terms of the derivatives of cot π‘₯ and sec π‘₯.

The main part of this question is finding the derivative of these two trigonometric functions: cot and sec. There are six trigonometric functions that we tend to use: sin, cos, tan, csc, sec, and cot. And certainly we could memorize all six of their derivatives if we wanted to, and this isn’t a terrible idea. It would allow us to differentiate any trigonometric function very quickly. Certainly if we memorized the derivatives of sec and cot, then we’d pretty much be done with this question.

But if you have forgotten the derivatives of sec and cot or if you’ve never memorized them in the first place, all is not lost. The key thing to do is to remember the first two derivatives: the derivative of sin π‘₯ with respect to π‘₯ is cos π‘₯, and the derivative of cos π‘₯ with respect to π‘₯ is minus sin π‘₯. From these two derivatives, we can derive the other four. Let’s see how we can find the derivative of cot when we know the derivatives of sin and cos.

Okay, so we’re looking for the derivative of cot π‘₯ with respect to π‘₯. Now what is cot π‘₯? Well, you might know that it’s one over tan π‘₯, and so we’re looking for the derivative of one over tan π‘₯ with respect to π‘₯. And as tan π‘₯ is sin π‘₯ over cos π‘₯, cot π‘₯ being the reciprocal of this must be cos π‘₯ over sin π‘₯. Now we’ve succeeded in writing cot π‘₯ in terms of sin and cos.

In fact, we’ve shown that cot π‘₯ is the quotient of cos π‘₯ and sin π‘₯. And so, we can differentiate it using the quotient rule. So here’s the quotient rule, and now we can apply it setting 𝑓 of π‘₯ to be cos π‘₯ and 𝑔 of π‘₯ to be sin π‘₯. Now we have the derivative we’re looking for written in terms of sin π‘₯ cos π‘₯ and their derivatives.

And we know the derivatives of sin π‘₯ and cos π‘₯, so we can substitute them in. The derivative of cos π‘₯ is minus sin π‘₯ and so the first term in the numerator becomes sin π‘₯ times minus sin π‘₯ or minus sin π‘₯ squared. How about the second term in the numerator? Well, the derivative of sin π‘₯ with respect to π‘₯ is cos π‘₯. And so, the second term becomes cos π‘₯ times cos π‘₯ or cos π‘₯ squared.

There’s not much we can do to the denominator. It stays as sin π‘₯ squared, which we can eventually write as sin squared π‘₯. But notice the numerator is minus sin squared π‘₯ minus cos squared π‘₯, which is minus sin squared π‘₯ plus cos squared π‘₯. And hopefully we recognize that we can apply the identity sin squared π‘₯ plus cos squared π‘₯ equals one, which means our numerator is negative one. And so the derivative of cot π‘₯ with respect to π‘₯ is negative one over sin squared π‘₯.

And taking the minus sign outside the fraction, we get negative one over sin squared π‘₯. We can apply this result to the derivative we wanted to find. The first term of 𝑑𝑦 by 𝑑π‘₯ is, therefore, eight times negative one over sin squared π‘₯. Before we move on, finding the derivative of sec π‘₯ with respect to π‘₯ and hence the second term of 𝑑𝑦 by 𝑑π‘₯, let’s first recap how we found the derivative of cot π‘₯ with respect to π‘₯.

We first wrote cot π‘₯ in terms of sin π‘₯ and cos π‘₯. And then seeing that we had a quotient, we applied the quotient rule using the derivatives of sin π‘₯ and cos π‘₯ with respect to π‘₯, which we memorized. And simplifying by applying the identity sin squared π‘₯ plus cos squared π‘₯ equals one gave us the simple expression negative one over sin squared π‘₯. So now let’s follow the same steps to find the derivative of sec π‘₯ with respect to π‘₯.

We can write sec π‘₯ in terms of sin π‘₯ and cos π‘₯. In fact, we only need cos π‘₯. Sec π‘₯ is one over cos π‘₯. And we differentiate this by applying the quotient rule with 𝑓 of π‘₯ equal to one and 𝑔 of π‘₯ equal to cos π‘₯. The derivative of one with respect to π‘₯ is just zero, and so the first term in the numerator vanishes. And so we’re left with just the second term: minus one times the derivative of cos π‘₯ with respect to π‘₯. And we know what the derivative of cos π‘₯ is with respect to π‘₯.

It’s minus sin π‘₯. So combining this with the minus one, we get sin π‘₯ in the numerator and of course cos squared π‘₯ in the denominator. And now that we found the derivative of sec π‘₯ with respect to π‘₯, we can apply this to our problem. The second term of 𝑑𝑦 by 𝑑π‘₯ becomes five sin π‘₯ over cos squared π‘₯. Okay, now we can clear away our working. Now we’ve found 𝑑𝑦 by 𝑑π‘₯ in terms of π‘₯. We now just need to evaluate this when π‘₯ is πœ‹ by six.

Now we replace π‘₯ by πœ‹ by six and we get some expression involving sin πœ‹ by six and cos πœ‹ by six. And as πœ‹ by six is a special angle, hopefully we know the values of sin πœ‹ by six and cos πœ‹ by six. Sin πœ‹ by six is a half and cos πœ‹ by six is root three over two. We can substitute these values and simplify. For example, a half squared is a fourth. And root three over two squared is three-fourths. Negative one over a fourth is negative four, and a half over three-fourths is two-thirds.

And so we get eight times negative four plus five times two-thirds. Simplifying, we find that the value of 𝑑𝑦 by 𝑑π‘₯ at π‘₯ equals πœ‹ by six is negative 86 over three. This problem required us to find the derivatives of cot and sec. We found that we could work out what these derivatives were using the known derivatives of sin and cos and also the quotient rule. The derivative of cot π‘₯ with respect to π‘₯ is [minus] one over sin squared π‘₯, which can also be written as minus csc squared π‘₯. And the derivative of sec π‘₯ with respect to π‘₯ we found to be sin π‘₯ over cos squared π‘₯, which can also be written as sec π‘₯ times tan π‘₯.

However, we left these derivatives in terms of sin and cos to make it easy to evaluate them at π‘₯ equals πœ‹ by six. We don’t need to memorize the derivatives of all six trigonometric functions, although it might be a good idea to do so as long as we remember the derivatives of sine and cos. To be fair, even the derivatives of sin and cos could be worked out from other more basic facts. But it’s definitely worth memorizing the derivatives of sin and cos at the very least.

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