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In this video, weβll learn how to differentiate the trigonometric functions sine, cosine, and tangent.
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Weβll begin by considering how we might find the derivative of the sine and cosine functions by using differentiation from first principles before using the quotient rule to find the derivative of the tangent function.
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Weβll then look at a few examples of the application of these derivatives and the patterns they form.
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By this point, you should feel comfortable in differentiating polynomial functions and applying the differentiation from first principles process.
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Remember, this says that the derivative of a function π is defined as π prime of π₯ is equal to the limit as β approaches zero of π of π₯ plus β minus π of π₯ all over β at points where the limit exists.
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So weβre going to use this definition to help us find the derivative of sin of π₯.
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But weβll also need to know some standard limits.
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It is possible to derive these, but for the purposes of this video, we will simply recall them.
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We must also note that for these limits to be true, we require any angles to be given in radiant measure.
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Weβre going to use that the limit as β approaches zero of sin of β over β is equal to one and the limit as β approaches zero of cos of β minus one over β is equal to zero.
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And now that weβve recalled the information required for this process, letβs see what it looks like.
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Differentiate π of π₯ equals sin π₯ from first principles.
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Weβre given that π of π₯ is equal to sin of π₯.
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And we know that to differentiate from first principles, we apply the formula the limit as β approaches zero of π of π₯ plus β minus π of π₯ all over β.
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So since π of π₯ is sin π₯, we need to define π of π₯ plus β.
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And thatβs sin of π₯ plus β.
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And so to differentiate sin π₯ from first principles, we apply the limit as β approaches zero of sin π₯ plus β minus sin of π₯ all over β.
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Notice that we canβt really evaluate anything yet.
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And so weβre going to need to find a way to simplify the expression sin π₯ plus β minus sin π₯ over β.
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And so we recall the sum formula for sin.
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This says that sin of π΄ plus π΅ is equal to sin π΄ cos π΅ plus cos π΄ sin π΅.
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And so we see that sin of π₯ plus β is the same as sin π₯ cos β plus cos π₯ sin β.
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And so weβre able to rewrite the limit representing our derivative of π of π₯ as shown.
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This still isnβt particularly useful, but what we can do is factor sin of π₯.
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We identify our two expressions that contain sin of π₯.
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And we see that sin π₯ cos β minus sin π₯ can now be written as sin of π₯ times cos of β minus one.
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Letβs just split this up a little.
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We separate the fraction and then recall that the limit of the sum of two functions is equal to the sum of their respective limits.
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And then we notice that sin of π₯ and cos of π₯ are actually independent of β.
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And so we can take both sin of π₯ and cos of π₯ outside of our limit.
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Next, we recall that for radiant measures, the limit as β approaches zero of sin of β over β is one.
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And the limit as β approaches zero of cos of β minus one over β is equal to zero.
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And so we see that we can replace one of our limits with zero and one of our limits with one.
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sin of π₯ times zero is zero.
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So this entire term disappears.
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cos of π₯ times one is simply equal to cos of π₯.
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And so weβve demonstrated by differentiation from first principles that π prime of π₯ is equal to cos of π₯.
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The derivative of sin π₯ is cos π₯.
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Now this is a result that should be learned by heart.
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But itβs also important that youβre able to follow the process for differentiating sin of π₯ by first principles.
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Letβs now look to repeat the process for cos of π₯.
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Given that π¦ is equal to cos of π₯, find dπ¦ by dπ₯ from first principles.
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To answer this, we can use the definition of a derivative.
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For some function π of π₯, the derivative which can be called π prime of π₯ but can be also defined as dπ¦ by dπ₯ is equal to the limit as β approaches zero of π of π₯ plus β minus π of π₯ over β where the limit exists.
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And β is equal to πΏπ₯.
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Now weβre going to let π of π₯ be equal to cos of π₯.
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So π of π₯ plus β is cos of π₯ plus β.
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And so the first line of working in our differentiation process is the limit as β approaches zero of cos of π₯ plus β minus cos of π₯ all over β.
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Then we recall the trigonometric identity that says that cos of π΄ plus π΅ is cos of π΄ times cos of π΅ minus sin of π΄ times sin of π΅.
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And so cos of π₯ plus β is cos π₯ cos β minus sin π₯ sin β.
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And so we can rewrite our limit as shown.
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Our next step is to factor cos of π₯.
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We notice we have cos of π₯ here and negative cos of π₯ over here.
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And so the numerator of this expression becomes cos of π₯ times cos β minus one minus sin of π₯ times sin β.
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Weβre going to split this into two fractions and apply the rules for working with limits.
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That is, the limit of the sum or difference of two functions is equal to the sum or difference of the respective limits.
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We also know that we can write, for instance, cos of π₯ times cos β minus one over β as cos of π₯ times the fraction cos β minus one over β.
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And so our limits look a little bit like this, but we know that cos of π₯ and sin of π₯ are independent of β.
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And so we take those outside of the limits.
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And so dπ¦ by dπ₯ is cos of π₯ times the limit as β approaches zero of cos β minus one over β minus sin of π₯ times the limit as β approaches zero of sin β over β.
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And then we apply the following limits, sometimes called the small angle approximation, noting again that this only works for radian measure.
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The limit as β approaches zero of cos β minus one over β is zero.
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So the first term becomes zero.
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Then negative sin π₯ times the limit as β approaches zero of sin β over one is negative sin π₯ times one which is just negative sin π₯.
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And so we see that for π¦ equals cos of π₯, dπ¦ by dπ₯, which is the first derivative of π¦ with respect to π₯, is negative sin π₯
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So for a real number π₯ given in radians, the first derivative of sin π₯ with respect to π₯ is cos π₯ and the first derivative of cos π₯ with respect to π₯ is negative sin π₯.
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Now, these actually form a pattern such that the derivative of sin π₯ is cos π₯.
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Differentiating again gives us negative sin π₯.
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Then when we differentiate that, we get negative cos π₯.
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And then one more derivative takes us back to sin π₯.
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Given that integration is the inverse of differentiation, we can also reverse this pattern when integrating.
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Now the formulae for derivatives of sine and cosine can actually be generalized to the derivatives of sin ππ₯ and cos ππ₯.
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The derivative of sin ππ₯ is π cos ππ₯ and the derivative of cos ππ₯ is negative π sin ππ₯.
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We can even consider the general formulas, and these hold for integer values of π.
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A similar set of formulae applies for the derivative of cos of π₯.
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So what about the derivative of the tangent function?
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Well, weβre going to apply an identity that relates tan of π to sin of π and cos of π.
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And weβre going to use the quotient rule.
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Letβs see what that looks like.
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Evaluate the rate of change of π of π₯ equals tan of five π₯ at π₯ equals π.
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Remember, when we think about the rate of change of a function, weβre really interested in its derivative.
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And so weβre actually going to differentiate tan of five π₯ and then evaluate that derivative when π₯ is equal to π.
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Weβll begin by rewriting tan of π₯, and weβre going to use the identity tan of π₯ equals sin of π₯ over cos of π₯.
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And so our π of π₯, which is tan of five π₯, can therefore be written as sin of five π₯ over cos of five π₯.
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And then to find the derivative of this rational function, weβre going to recall the quotient rule.
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This says that the derivative with respect to π₯ of the quotient of two differentiable functions π’ over π£ is given by π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ all over π£ squared.
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And so to use this, we let π’ be equal to sin of five π₯, thatβs the numerator, and π£ be equal to cos of five π₯; thatβs the denominator.
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We will, of course, need to find the derivative of each of these functions with respect to π₯.
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And, of course, we know that the derivative of sin ππ₯ is π cos ππ₯.
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So the derivative of sin of five π₯ must be five cos of five π₯.
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Then when we differentiate cos of ππ₯ with respect to π₯, we get negative π sin of ππ₯.
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And so the derivative of π£ with respect to π₯ is negative five sin of five π₯.
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So we substitute all of this into our formula for the quotient rule.
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We have π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ all over π£ squared.
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But then we see that we can rewrite the numerator as five cos squared five π₯ plus five sin squared five π₯.
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And then we factor the five from the numerator.
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And thatβs really useful because we have a trigonometric identity we can use.
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We know that cos squared π₯ plus sin squared π₯ is equal to one.
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And so cos squared five π₯ plus sin squared five π₯ must also be equal to one.
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So this becomes five over cos squared five π₯.
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But we use yet another identity to rewrite this.
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One over cos π₯ is sec π₯, so one over cos squared π₯ must also be sec squared π₯.
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And so five over cos squared five π₯ can be written as five sec squared five π₯.
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And so that is the derivative of tan of five π₯.
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Now, remember, we were wanting to evaluate this at π₯ equals π.
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So π prime of π is five sec squared five π.
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But sec squared five π is one.
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So we simply get five.
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And so the rate of change of π of π₯ equals tan five π₯ at π₯ equals π is five.
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Now, in fact, we can generalize the result here from differentiating tan of five π₯.
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The derivative of tan π₯ with respect to π₯ is sec squared π₯.
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And the derivative of tan ππ₯ for real constants π with respect to π₯ is π sec squared ππ₯.
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Now, of course, when we differentiated sin π₯ and cos π₯, we did say those results hold for radian measure.
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And so since we use those results to differentiate tan of five π₯, we know that this only applies for a radian measure too.
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Now, as with the derivatives of sine and cosine, itβs important to know this result by heart, but also be prepared to derive it when necessary.
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Weβre now going to have a look at an example of the application of these derivatives.
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If π¦ is equal to π₯ to the fifth power times sin of five π₯, determine dπ¦ by dπ₯.
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Here, we have a function which is the product itself of two differentiable functions.
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And weβre therefore going to use the product rule to differentiate it.
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This says that if π’ and π£ are differentiable functions, the derivative of their product is π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯.
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And so we begin by defining π’ and π£.
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Now multiplication is commutative, which means it doesnβt really matter which way round we define these.
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Letβs let π’ be equal to π₯ to the fifth power and π£ be equal to sin five π₯.
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Then dπ’ by dπ₯, the derivative of π₯ to the fifth power with respect to π₯ is five π₯ to the fourth power.
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And since we know how to differentiate sin of ππ₯, we get π cos ππ₯.
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The derivative of sin of five π₯ is five cos of five π₯.
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And so we substitute this into the formula.
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And we get π’ times dπ£ by dπ₯ to be equal to π₯ to the fifth power times five cos five π₯ plus π£ times dπ’ by dπ₯.
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And thatβs sin five π₯ times five π₯ to the fourth power.
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And so we see dπ¦ by dπ₯ is five π₯ to the fifth power cos five π₯ plus five π₯ to the fourth power sin five π₯.
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Weβll consider one final example which involves a little bit of manipulation.
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If π¦ equals two sin of seven π₯ plus two cos of seven π₯ squared, find dπ¦ by dπ₯.
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There are a number of ways to answer this question.
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We might, for example, spot that π¦ is a composite function and use the chain rule.
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We could also write it as the product of two functions and use the product rule.
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Alternatively, we can actually simplify it if we use some trigonometric identities.
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Letβs just begin by factoring two out of the parentheses, remembering, of course, we need to square it in the process.
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Next, letβs distribute our parentheses.
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sin seven π₯ plus cos seven π₯ squared is the same as sin seven π₯ plus cos seven π₯ times sin seven π₯ plus cos seven π₯.
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And that is equal to sin squared seven π₯ plus two sin seven π₯ cos seven π₯ plus cos squared seven π₯.
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Of course, two squared is also equal to four.
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Next, we use the trigonometric identity sin squared π₯ plus cos squared π₯ equals one.
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And we apply it to the sum of sin squared seven π₯ and cos squared seven π₯.
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Thatβs also one.
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And so we rewrite the expression inside our parentheses to get one plus two sin of seven π₯ cos of seven π₯.
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Do you spot another identity here?
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We can actually use the inverse of the double angle identity.
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sin of two π is two sin π cos π.
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And so weβre going to let π be equal to seven π₯.
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And that tells us that two sin of seven π₯ cos of seven π₯ is equal to sin of two times seven π₯, which is sin of 14π₯.
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We can now differentiate this fairly easily.
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Weβll use the fact that the derivative with respect to π₯ of sin ππ₯ is π cos ππ₯.
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And weβre going to use the fact that we can take constants outside of a derivative and concentrate on differentiating the function of π₯ itself.
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So dπ¦ by dπ₯ will be four times the derivative of one plus sin of 14π₯.
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The derivative of one with respect to π₯ is zero and then the derivative of sin of 14π₯ is 14 cos of 14π₯.
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And so we end up multiplying four by 14 cos of 14π₯ to give us 56 cos of 14π₯.
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If π¦ is the function given then, dπ¦ by dπ₯ is 56 cos of 14π₯.
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In this video, we learned that we can apply our understanding of derivatives to trigonometric functions.
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We saw that the derivative of sin ππ₯ with respect to π₯ is π cos ππ₯.
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We saw that the derivative of cos ππ₯ was negative π sin ππ₯.
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And the derivative of tan ππ₯ was π sec squared ππ₯.
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Because of the nature of how these were derived, though, we saw that they only apply when angles are given in radians.
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We also saw that the derivative of sine and cosine perform a pattern.
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And we can use this pattern to differentiate and integrate sin of π₯ or cos of π₯ a number of times.