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In this video, weβll learn how to find the derivatives of the inverses of trigonometric functions.
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We will learn how to do this by using implicit differentiation.
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And thus, it is important that you understand how to apply the chain rule, if not have a thorough understanding of how implicit differentiation works, before watching this video.
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After deriving the derivatives of the inverse trigonometric functions, weβll consider the application of these derivatives to more complicated inverse trigonometric functions.
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Before we look at finding the derivative of our inverse trigonometric functions, letβs just quickly consider the function π¦ equal sin of π₯.
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Remember, π₯ is a real number.
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And of course, since weβre performing calculus with a trigonometric function, we need to make sure this is measured in radians.
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For this function, we can say that π₯ is equal to the inverse sin of π¦.
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This subscript negative one denotes the inverse function.
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Remember though without restricting the domain of arc sin of π₯ or inverse sin of π₯, the function is going to be many to one.
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We, therefore, restrict the domain for π of π₯ equals inverse sin of π₯.
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And we say that π₯ has to be greater than or equal to negative one and less than or equal to one.
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If we go back to the function π₯ equals the inverse sin of π¦, we can, therefore, see that π₯ will take values greater than or equal to negative π by two and less than or equal to π by two.
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Weβll also need to recall the chain rule.
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This says that if π¦ is some function in π’ and π’ is some differentiable function in π₯, then dπ¦ by dπ₯ is equal to dπ¦ by dπ’ times dπ’ by dπ₯.
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Weβll now use everything weβve seen here to find the derivative of the inverse sine function.
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Find the derivative of the inverse sine function with respect to π₯.
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Weβre differentiating the inverse sin of π₯ with respect to π₯.
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So weβre going to begin by letting π¦ be equal to the inverse sin of π₯.
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Then we can say that π₯ must be equal to sin of π¦.
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Weβre going to differentiate both sides of this equation with respect to π₯.
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So we say that d by dπ₯ of π₯ is equal to d by dπ₯ of sin of π¦.
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Well, the derivative of π₯ with respect to π₯ is quite straight forward; itβs one.
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But weβre going to need to use implicit differentiation which is a special case of the chain rule to differentiate sin π¦ with respect to π₯.
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The derivative of sin π¦ with respect to π¦ is cos π¦.
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So the derivative of sin π¦ with respect to π₯ is cos π¦ times the derivative of π¦ with respect to π₯ which is just dπ¦ by dπ₯.
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So we currently see that one is equal to cos of π¦ times dπ¦ by dπ₯.
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We divide both sides of this equation by cos π¦ to form an equation for the derivative.
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And we see that dπ¦ by dπ₯ is equal to one over cos π¦.
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Now, weβve got a bit of a problem.
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We do want an expression for the derivative in terms of π₯ not π¦.
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And remember, we said that π₯ was equal to sin of π¦.
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So weβll use the identity cos squared π plus sin squared π equals one.
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And Iβve replaced π with π¦.
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Weβll subtract sin squared π¦ from both sides of the equation.
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And then weβll take the square root of both sides.
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And we see that cos of π¦ is equal to the positive and negative square root of one minus sin squared π¦.
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Remember, inverse sin is restricted to the closed interval negative π by two to π by two.
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By our definition, that means π¦ must be greater than or equal to π two and less than or equal to π by two which, in turn, means cos of π¦ must be greater than or equal to zero and less than or equal to one.
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And thatβs because in the interval π¦ is greater than or equal to negative π by two and less than or equal to π by two.
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The smallest value cos of π¦ weβll take is zero.
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And the greatest value is one.
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And what this means here is weβre going to take the positive square root of one minus sin squared π¦ only.
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We can now replace sin of π¦ with π₯.
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And we see that cos of π¦ is equal to the square root of one minus π₯ squared.
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And, therefore, dπ¦ by dπ₯ equals one over the square root of one minus π₯ squared.
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And we found the derivative of the inverse sin of π₯.
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Itβs one over the square root of one minus π₯ squared for values of π₯ in the range π₯ is greater than negative one and less than one.
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In our next example, weβll consider an alternative method that will help us find the derivative of the inverse cosine function.
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This time, weβre going to need to know the inverse function theorem.
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This says that if π is a differentiable function with a continuous inverse π prime and π prime of π is not equal zero, then not only is π invertible but it has a differentiable inverse.
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Such that the derivative of the inverse of π at some π equals π of π is equal to one over the derivative of π at π.
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This is sometimes written simply as dπ₯ by dπ¦ is equal to one over dπ¦ by dπ₯.
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Letβs see how this might help us when differentiating the inverse cosine function.
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Find the derivative of the inverse cos of π₯ over π with respect to π₯, where π is not equal to zero.
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Weβll begin by letting π¦ be equal to the inverse cos of π₯ over π.
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This can be alternately written as π₯ over π equals cos of π¦.
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And we can then multiply both sides by π.
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And we see that π₯ is equal to π times cos of π¦.
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Weβre going to differentiate our expression for π₯ with respect to π¦.
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In other words, weβre going to find dπ₯ by dπ¦.
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Weβll use the general result that the derivative of cos of π₯ with respect to π₯ is negative sin of π₯.
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And we see that dπ₯ by dπ¦ must be equal to negative π sin of π¦.
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Now, before we perform the next step we need to recall the fact that for the inverse trigonometric functions, we restrict their domains.
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And we know that the domain of the inverse cos of π₯ or our cos of π₯ is greater than or equal to zero and less than or equal to π.
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This means that π¦ must be greater than or equal to zero and less than or equal to π.
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Now, weβre going to use the inverse function theorem.
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So weβre going to take values of π¦ greater than zero and less than π, such that sin of π¦ is not equal to zero.
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Using this criteria, we can use dπ₯ by dπ¦ equals one over dπ¦ by dπ₯ which can be rearranged to say that dπ¦ by dπ₯ equals one over dπ₯ by dπ¦.
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And we see that, for our case, dπ¦ by dπ₯ equals one over negative π sin of π¦.
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And we have an expression for the derivative in terms of π¦.
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Remember, we want this to be in terms of π₯.
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We said that π₯ over π equals cos of π¦.
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So weβll use the fact that sin squared π¦ plus cos squared π¦ equals one and rearrange this to say that sin π¦ equals plus or minus the square root of one minus cos squared π¦.
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When π¦ is between zero and π, sin π¦ is greater than zero.
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So in fact, weβre only interested in the positive root.
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So weβll replace this in our expression for the derivative.
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And we get negative one over π times the square root of one minus cos squared π¦.
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We then replace cos π¦ with π₯ over π and change π₯ over π all squared to π₯ squared over π squared.
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And then we bring π into the square root.
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And we see that dπ¦ by dπ₯ is equal to negative one over the square root of π squared minus π₯ squared.
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So d by dπ₯ of the inverse cos of π₯ over π is equal to negative one over the square root of π squared minus π₯ squared for values of π₯ between negative π and π.
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In our next example, weβll consider how we might apply the process thatβs used so far to find the derivative of the inverse tangent function.
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Find an expression for the derivative of π¦ equals the inverse tan of ππ₯ in terms of π₯.
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Since π¦ is equal to the inverse tan of ππ₯, we can write ππ₯ as being equal to tan π¦.
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Weβre going to use implicit differentiation to find the derivative of both sides of this equation.
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The derivative of ππ₯ with respect to π₯ is simply π.
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And the derivative of tan π¦ with respect to π₯ is equal to the derivative of tan π¦ with respect to π¦ times the derivative of π¦ with respect to π₯.
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The derivative of tan π₯ is sec squared π₯.
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And the derivative of π¦ with respect to π₯ is dπ¦ by dπ₯.
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So we see that π is equal to sec squared π¦ times dπ¦ by dπ₯.
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Dividing through by sec squared π¦ and we see that dπ¦ by dπ₯ equals π over sec squared π¦.
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Weβre going to need to represent our equation for the derivative in terms of π₯.
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So weβll use this trigonometric identity.
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One plus tan squared π₯ equals sec squared π₯.
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This means we can write dπ¦ by dπ₯ as π over one plus tan squared π¦.
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And then we replaced tan π¦ with ππ₯.
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And we see the expression for the derivative of π¦ equals the inverse tan of ππ₯ is π over one plus ππ₯ squared.
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Similar rules can be applied to help us find the derivative of the inverse cotangent function.
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We find that the derivative of the inverse of cotangent of π of π₯ is equal to negative π over one plus ππ₯ squared.
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The inverse cosecant and secant functions are a little more unusual.
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So weβll consider next how to find the derivative of the inverse cosecant function.
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Find d by dπ₯ of the inverse cosecant of π₯.
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We begin by letting π¦ be equal to the inverse cosecant of π₯.
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And this means we can rewrite this.
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And we can say that π₯ is equal to the cosecant of π¦.
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Weβre next going to use implicit differentiation to find the derivative of both sides of this equation.
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The derivative of π₯ with respect to π₯ is simply one.
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Then the derivative of cosec π¦ with respect to π₯ is equal to the derivative of cosec π¦ with respect to π¦ times dπ¦ by dπ₯.
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And the derivative of cosec π¦ with respect to π¦ is negative cosec π¦ cot π¦.
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So we see that one is equal to negative cosec π¦ cot π¦ times dπ¦ by dπ₯.
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Now, we know that, for the inverse cosecant function, π¦ must be greater than negative π by two and less than π by two and not equal to zero.
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Using these restrictions cosec π¦ cot π¦ cannot be equal to zero.
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So we can divide through by negative cosec π¦ cot π¦.
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And we see that dπ¦ by dπ₯ is as shown.
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We want to represent our equation for the derivative in terms of π₯.
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So weβll use this trigonometric identity cot squared π¦ plus one equals cosec squared π¦.
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And we can rewrite this to say that cot of π¦ is equal to the positive and negative square root of cosec squared π¦ minus one.
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We put this into the equation for the derivative in place of cot of π¦.
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And we then use the fact that π₯ is equal to cosec π¦.
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But we are going to need to make a decision on the sine of the derivative.
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And it can help here to look at the graph of the inverse cosecant function.
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Notice how, for all values of π₯ in the range of the function, the derivative of the slope of the tangent is negative.
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And we, therefore, use the absolute value to ensure that our derivative is always negative.
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We say that dπ¦ by dπ₯ is equal to the negative of the absolute value of one over π₯ times the square root of π₯ squared minus one.
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Since one and the square root of π₯ squared minus one are always positive, we can rewrite this as shown.
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So the derivative of the inverse cosecant function π₯ with respect to π₯ is negative one over the modulus or absolute value of π₯ times the square root of π₯ squared minus one.
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A similar process can be applied to help us find the derivative of the inverse secant function.
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And we have the derivatives of all of the inverse trigonometric functions we require.
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Itβs useful to commit these results to memory but also be prepared to derive them where necessary.
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Weβll now have a look at the application of these results.
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Evaluate the derivative of the inverse cotangent of one over π₯ with respect to π₯.
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Here, we have a function of a function or a composite function.
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Weβre, therefore, going to need to use the chain rule to find the derivative.
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This says that if π and π are differentiable functions such that π¦ is π of π’ and π’ is π of π₯, then dπ¦ by dπ₯ is equal to dπ¦ by dπ’ times dπ’ by dπ₯.
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Weβll let π’ be equal to one over π₯.
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Then π¦ is equal to the inverse cot of π’.
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To apply the chain rule, we need to find the derivative of both of these functions.
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And with π’ it can be useful to write it as π₯ to the negative one.
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Then dπ’ by dπ₯ is negative π₯ to negative two or negative one over π₯ squared.
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We can then use the general derivative of the inverse cotangent function.
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And we see that dπ¦ by dπ’ is equal to negative one over one plus π’ squared. dπ¦ by dπ₯ is the product of these.
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Itβs negative one over π₯ squared times negative one over one plus π’ squared.
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We can replace π’ with one over π₯ and then multiply through.
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And we see that the derivative of the inverse cotangent of one over π₯ with respect to π₯ is one over π₯ squared plus one.
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Did you notice that the derivative of the inverse cotangent of one over π₯ is equal to the derivative of the tangent of π₯?
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This is, in fact, no accident.
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And we can use the identity the inverse cotangent of one over π₯ equals the inverse tan of π₯.
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That could have saved us a little bit more time in this previous example.
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Evaluate the derivative of the inverse sin of the square root of one minus π₯ squared with respect to π₯.
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Here, we have a function of a function or a composite function.
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So weβll use the chain rule to find its derivative.
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This says that if π¦ is some function in π’ and π’ is some function in π₯, then dπ¦ by dπ₯ is equal to dπ¦ by dπ’ times dπ’ by dπ₯.
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Weβll let π’ be equal to the square root of one minus π₯ squared.
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Which can, of course, alternatively be written as one minus π₯ squared to the power of one-half.
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Then π¦ is equal to the inverse sin of π’.
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To apply the chain rule, weβre going to need to find the derivative of both of these functions.
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The derivative of the inverse sin of π’ with respect to π’ is one over the square root of one minus π’ squared.
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And we can use the general power rule to find the derivative of one minus π₯ squared to the power of one-half.
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Itβs a half times one minus π₯ squared to the negative one-half times the derivative of the bit inside the brackets which is negative two π₯.
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That can be written as negative π₯ times one minus π₯ squared to the power of negative one-half.
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dπ¦ by dπ₯ is, therefore, negative π₯ over the square root of one minus π₯ squared times one over the square root of one minus π’ squared.
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We can replace π’ with one minus π₯ squared to the power of one-half.
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And the second fraction becomes one over the square root of one minus one minus π₯ squared.
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This further simplifies to one over π₯.
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And we divide through by π₯.
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And we see that the derivative of our function is negative one over the square root of one minus π₯ squared.
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Once again, weβve stumbled across an interesting result.
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That is that the derivative of the inverse sin of the square root of one minus π₯ squared is equal to the derivative of the inverse cos of π₯.
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This comes from the identity of the inverse sin of the square root of one minus π₯ squared is equal to the inverse cos of π₯, the values of π₯ between zero and one.
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Being familiar with this result could have reduced the amount of what we needed to do in this example.
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In this video, weβve seen that we can use implicit differentiation or the inverse function theorem to derive the formulae for the derivatives of inverse trigonometric functions.
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We saw that the derivatives of the inverse trigonometric functions are as shown.
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And we also saw that being familiar with certain trigonometric identities can sometimes significantly reduce the process of finding these derivatives.