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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 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 that they form.
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By this stage, you should feel comfortable differentiating polynomial functions and applying the differentiation from first principles process.
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Remember, this says that we can find the derivative of a function π by using the following formula.
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Itβs the limit as β tends to zero of π of π₯ plus β minus π of π₯ all over β.
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And thatβs at the points where the limit exists.
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Weβre going to begin by using this definition to help us find the derivative of sin of π₯.
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Weβre also going to need to know some standard limits.
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Now whilst it is possible to derive these, for the purposes of this video, weβre simply going to recall them.
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And whilst we consider these, we must note that, for them to be true, we require any angles to be given in radians.
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These say that the limit as β tends to zero of sin β over β is one.
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And the limit as β tends to zero of cos of β minus one over β is zero.
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Now that weβve recalled the information required for this process, letβs see what it looks like.
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Differentiate π of π₯ is equal to sin π₯ from first principles.
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To differentiate from first principles, we use the formula π prime of π₯ is equal to the limit as β tends to zero of π of π₯ plus β minus π of π₯ all over β.
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Weβve already been told that π of π₯ is equal to sin π₯.
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And this means then that π of π₯ plus β which is what we need for our formula is sin of π₯ plus β.
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We can now write the derivative of our function in π₯ as the limit is β tends to zero of sin of π₯ plus β minus sin of π₯ all over β.
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Notice at this point, we canβt actually evaluate anything yet.
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So weβre going to need to simplify the expression sin of π₯ plus β minus sin of π₯ over β.
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And we do this by recalling the sum formula for sine.
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That tells us that sin of π΄ plus π΅ is equal to sin of π΄ times cos of π΅ minus cos of π΄ time sin of π΅.
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We can, therefore, say that sin of π₯ plus β is equal to sin π₯ cos β minus cos π₯ sin β.
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And this means weβre going to be evaluating sin π₯ cos β minus cos π₯ sin β minus sin π₯ over β as β tends to zero.
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But we still canβt evaluate this.
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So weβre going to manipulate the expression somewhat.
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To do this we factor sin π₯.
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And we get sin π₯ of cos β minus one minus cos π₯ sin β over β.
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Now, we have these two standard results.
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The limit as β tends to zero of sin β over β is equal to one.
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And as β tends to zero for the expression cos β minus one over β, we get zero.
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And remember, these are for radian values.
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And now, we look to split our expression up.
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We get sin π₯ cos β minus one over β minus cos π₯ sin β over β.
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And of course, sin π₯ and cos of π₯ are independent of β.
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And this means that sin π₯ times cos β minus one over β as β tends to zero tends to sin π₯ time zero.
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And as β tends to zero cos of π₯ times sin β over β tends to cos of π₯ times one.
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So we see that the derivative of sin π₯ with respect to π₯ is sin π₯ times zero plus cos π₯ times one.
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Well, thatβs simply cos of π₯.
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So we see the π prime of π₯ or the derivative of sin π₯ is cos of π₯.
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Now, this is a result that should be learnt by heart.
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But itβs also important that youβre able to follow the process to differentiate sin π₯ by first principles.
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Weβre going to look to repeat this 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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Given that the formula for differentiating by first principles is in function notation, we begin by letting π of π₯ be equal to cos of π₯.
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And this means that π of π₯ plus β is equal to cos of π₯ plus β.
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And then we use that definition of a derivative and substitute these values into our function.
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And we get the limit as β tends to zero of cos of π₯ plus β minus cos of π₯ over β as the derivative.
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Now, we canβt evaluate this yet.
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So we use the identity cos of π΄ plus π΅ is equal to cos π΄ cos π΅ minus sin π΄ sin π΅.
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And that means we can write cos of π₯ plus β as cos π₯ cos β minus sin π₯ sin β.
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And we see that the derivative of our function is given by the limit as β tends to zero of cos π₯ cos β minus sin π₯ sin β minus cos π₯ over β.
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And then we noticed that we have a repeated factor.
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We can factor cos of π₯.
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And we get cos of π₯ times cos β minus one minus sin π₯ sin β all over β.
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And then we recall the fact that, for radium values, the limit as β tends to zero of sin β over β is one.
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And the limit as β tends to zero of cos β minus one over β is zero.
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And this prompts us to split the expression up a little bit.
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We write it as cos π₯ times cos β minus one over β minus sin π₯ times sin β over β.
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And remember, sin π₯ and cos of π₯ are independent of β.
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And we can then see that the limit as β tends to zero of cos π₯ times cos β minus one over β is cos π₯ times zero.
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And as β tends to zero, sin π₯ times sin β over β tends to sin π₯ times one.
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And so π prime of π₯ is cos of π₯ times zero minus sin of π₯ times one which if we revert back to limits notation shows us that the derivative dπ¦ by dπ₯ is equal to negative sin π₯ for radium values of π₯.
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These previous two examples have shown us that, for a real number π₯ given in radians, the derivative of sin π₯ with respect to π₯ is cos of π₯.
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And the derivative of cos of 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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The derivative of cos π₯ is negative sin π₯.
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When we differentiate negative sin π₯, we get negative cos π₯.
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And when we differentiate negative cos π₯, we go back to sin π₯.
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And of course, since integration is the opposite of differentiation, we can reverse this pattern when integrating.
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Generalising this for higher-order derivatives, we see that for integer values of π the general formulas for sin of π₯ hold.
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And we have similar formulas for the derivative of cos of π₯.
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And we can generalize these.
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We could look at how this works using the chain rule.
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But actually just for the purposes of this video, weβre going to state the derivatives.
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The derivative of sin ππ₯ with respect to π₯ is equal to π cos of ππ₯.
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And the derivative with respect to π₯ of cos ππ₯ is negative π sin ππ₯.
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But what about the derivative of the tangent function?
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Well, the derivative of the tangent function relies on two pieces of information.
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The first is the identity tan π equals sin π over cos π.
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The second is the quotient rule.
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And this is that, given two differentiable functions, π’ and π£, the derivative of a quotient π’ over π£ is equal to π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ all over π£ squared.
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Letβs apply these in an example.
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Evaluate the rate of change of π of π₯ is equal to tan five π₯ at π₯ equals π.
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Remember, when considering the rate of change of a function, weβre really interested in finding its derivative.
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Weβre, therefore, going to need to differentiate tan of five π₯ with respect to π₯ and then evaluate the derivative at π₯ equals π.
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Weβre going to begin then by rewriting tan of five π₯, using the identity tan π equals sin π over cos π.
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And this means tan of five π₯ is equal to sin five π₯ over cos five π₯.
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Letβs use the quotient rule to evaluate the derivative.
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Since the numerator of our fraction is sin five π₯, we let π’ be equal to sin five π₯.
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And we let π£ be equal to cos of five π₯.
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And this means the derivative of π’, the derivative of sin five π₯, with respect to π₯ is five cos five π₯.
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And since the derivative of cos of ππ₯ with respect to π₯ is equal to negative π sin ππ₯, dπ£ by dπ₯ is negative five sin five π₯.
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Letβs substitute these into the formula.
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π£ times dπ’ by dπ₯ is cos five π₯ times five cos five π₯.
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And π’ times dπ£ by dπ₯ is sin five π₯ times negative five sin five π₯.
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And of course, this is all over π£ squared which we can say is cos squared five π₯.
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Simplifying the numerator gives us five cos squared five π₯ plus five sin squared five π₯.
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And then we factor five on the numerator to get five times cos squared five π₯ plus sin squared five π₯.
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And we do this because we know that sin squared π plus cos squared π is equal to one.
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And so π prime of π₯ is five over cos squared five π₯.
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But thereβs yet another identity we can use.
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One over cos squared π [cos π] is the same as sec squared π [sec π], which means that five over cos squared five π₯ is going to be the same as five sec squared five π₯.
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Remember, weβre looking to find the rate of change at π₯ equals π.
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So we substitute π into our formula for π prime of π₯.
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And we get π prime of π is five sec squared five π which is just five.
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In fact, the general result of the derivative of tan π₯ with respect to π₯ is sec squared π₯.
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And the derivative of tan ππ₯ is π sec squared ππ₯.
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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 a number of examples of the application of these derivatives.
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If π¦ is equal to π₯ to the power of five sin five π₯, determine dπ¦ by dπ₯.
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Here, we have a function which is the product of two differentiable functions.
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We can, therefore, use the product rule to help us evaluate dπ¦ by dπ₯.
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This says that, for our function π¦ is equal to π’ times π£, the derivative of π¦ with respect to π₯ is equal to π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯.
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Weβre going to let π’ be equal to π₯ to the power of five and π£ be equal to sin five π₯.
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We begin by evaluating the derivative of π’ with respect to π₯.
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Itβs five π₯ to the power of four.
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Similarly, since the derivative of sin ππ₯ is π cos ππ₯, dπ£ by dπ₯ is equal to five cos five π₯.
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Then we substitute into the formula.
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π’ times dπ£ by dπ₯ is π₯ to the power of five times five cos five π₯.
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And π£ times dπ’ by dπ₯ is sin five π₯ times five π₯ to the power of four.
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And so dπ¦ by dπ₯, in this case, is five π₯ to the power of five times cos five π₯ plus five π₯ to the power of four times sin five π₯.
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Weβre going to consider one final example which involves some manipulation of the expression given to us.
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If π¦ is equal to two sin seven π₯ plus two cos 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 could spot that π¦ is a composite function.
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Itβs the square of two sin seven π₯ plus two cos seven π₯.
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And we could use the chain rule to evaluate its derivative.
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We could also write it as a product of two functions and use the product rule.
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Alternatively though, we can simplify this by using known trigonometric identities.
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Weβre going to use that third method.
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And weβll begin by factoring two out of the parentheses.
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That gives us π¦ is equal to two squared times sin seven π₯ plus cos seven π₯ squared.
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Next, we look to distribute the parentheses.
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Multiplying the first term in each bracket gives us sin squared seven π₯.
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Multiplying the outer terms gives us sin seven π₯ times cos seven π₯.
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Multiplying the inner terms gives us cos seven π₯ sin seven π₯.
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And then we multiply the last two terms in each bracket.
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And we get cos squared seven π₯.
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Now, we can actually simplify this a little bit.
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We know that sin squared π₯ plus cos squared π₯ is equal to one.
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Well that means that sin squared seven π₯ plus cos squared seven π₯ must also be equal to one.
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And so we replace this with one and we collect like terms.
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And we see that π¦ is equal to four times one plus two sin seven π₯ cos seven π₯.
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Can you spot another identity here?
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We need to use the inverse of the double angle formula for sin.
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Itβs sin two π΄ equals two sin π΄ cos π΄.
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And weβre going to let π΄ be equal to seven π₯.
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And we see that π¦ is equal to four times one plus sin 14π₯ or four plus four sin 14π₯.
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And we can now differentiate this fairly easily.
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The derivative of sin ππ₯ with respect to π₯ is π cos ππ₯.
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We also use the constant multiple rule which allows us to take constants outside the derivative and concentrate on differentiating the function of π₯ itself.
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The derivative of four is zero.
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So dπ¦ by dπ₯ is equal to four times 14 cos of 14π₯ which is 56 cos of 14π₯.
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In this video, weβve learnt that we can apply our understanding of derivatives to trigonometric functions.
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We also learn that calculus relating to trigonometric functions can be applied when angles are given in radians.
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And we saw that the derivatives of the sine, cosine, and tangent functions are the derivative of sin ππ₯ is π cos ππ₯, the derivative of cos ππ₯ is negative π sin ππ₯, and the derivative of tan ππ₯ is π sec squared ππ₯.