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Differentiation of Inverse Functions
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In this video, we will learn how to find the derivative of inverse functions.
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And weβll be covering a variety of examples of how we can do so.
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Letβs start by recapping some information about inverse functions.
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Let π be a function with domain π and range π, so π takes π to π.
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We call the function π, which takes π to π, the inverse of π, if for all π¦ contained in π, π of π of π¦ is equal to π¦.
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And for all π₯ contained in π, π of π of π₯ is equal to π₯.
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What this definition is telling us is that if some function π has an inverse and then we apply both π and the inverse to some value.
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Then we will obtain the value itself.
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Now, if the function π from this definition exists, then we can say that π is invertible and π is the inverse of π.
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We can also denote the inverse function like this with π superscript negative one.
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However, we must be careful not to confuse this with π to the power of negative one.
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Since they are completely different despite similar notation.
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If there is a superscript negative one next to a function, it means the inverse.
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But if itβs next to a variable or a constant, it means to the power of negative one.
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Another thing to note is that if π is the inverse of π, then π is also the inverse of π.
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Letβs now consider the graph of some function π.
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Now, we know that we can find the graph of the inverse of this function by reflecting it in the line π¦ equals π₯, which is this line here.
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So our inverse function will look something like this.
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What weβre aiming to do here is to find the derivative of the inverse function.
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Now, the derivative is the function of the slope of the graph.
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And one way to find the slope at a given point is to find the tangent at that point and then find the slope of the tangent.
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Letβs find the tangent to π at some π₯-value, which we can call π.
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Now, the coordinates of the point where we are taking the tangent will be π π of π.
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However, we can also call π of π π such that the point at which weβre taking the tangent is at π, π.
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Now, this gives us that π of π is equal to π.
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Now, we somehow need to link this point to the inverse function of π.
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If we apply π inverse to both sides here, we will obtain that π inverse of π of π is equal to π inverse of π.
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However, due to the definition of an inverse function, we know that π inverse of π of π is equal to π.
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This gives us that π inverse of π is equal to π.
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Letβs find this point on π inverse of π₯.
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And we can find the tangent to π inverse of π₯ at this point.
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Now, it looks as though our tangents of π of π₯ at π, π and π inverse of π₯ at π, π are reflections of one another in the line π¦ equals π₯.
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And this would make sense, since π of π₯ is a reflection of π inverse of π₯ in the line π¦ equals π₯.
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And the point π, π is the reflection of the point π, π in the line π¦ equals π₯.
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Now, what weβre concerned with here is this slope function of π inverse of π₯.
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So letβs consider the slope of these two tangents.
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Letβs call the tangent of π of π₯ πΏ one and the tangent of π inverse of π₯ πΏ two.
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We can call the equation of πΏ one π¦ is equal to ππ₯ plus π.
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Now, a reflection in the line π¦ equals π₯ corresponds to a mapping of π₯, π¦ going to π¦, π₯.
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Therefore, the equation of line πΏ two is, π₯ is equal to ππ¦ plus π.
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And we can rearrange this equation to make π¦ the subject, giving us that π¦ is equal to one over ππ₯ minus π over π.
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Another thing which weβre concerned with here is the slopes of these two tangents.
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We can see that the slope of the tangent to π of π₯ is π.
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And the slope of the tangent to π inverse of π₯ is one over π.
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Since π is the slope of the tangent at π, π, it can also be defined as the slope of π at π₯ equals π.
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And so, we have that π is equal to π prime of π.
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One over π represents the slope of the tangent to π inverse of π₯ at π, π.
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So itβs the slope of π inverse at π₯ equals π.
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So we can say that the derivative of π inverse at π is equal to one over π.
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However, we just found out that π is equal to π prime of π.
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So this can be substituted in.
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And this gives us our result.
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And that is that if π of π is equal to π, then the derivative of the inverse of π at π is equal to one over the derivative of π at π.
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Which, of course, only makes sense if π prime of π is nonzero.
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Now, we havenβt proven this result rigorously.
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Since we have only based the fact that the tangents are reflections of one another off of intuition.
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This result can, however, be proven using the chain rule.
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If we have some function π with an inverse function π, then by the definition of the inverse function, π of π of π¦ is equal to π¦.
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Now, if we use the chain rule to differentiate both sides of this equation with respect to π¦.
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Then we will obtain that π prime of π of π¦ multiplied by π prime of π¦ is equal to one.
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Rearranging, we obtain our result.
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Which is that π prime of π¦ is equal to one over π prime of π of π¦.
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This is quite often also written in Leibnizβs notation.
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Which tells us that dπ¦ by dπ₯ is equal to one over dπ₯ by dπ¦.
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Letβs now apply these definitions to some examples.
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Given that π₯ is equal to π to the power of π¦, find dπ¦ by dπ₯, giving your answer in terms of π₯.
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We can start by differentiating π₯ with respect to π¦.
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Using the fact that the differential of an exponential is just the exponential, we obtain that dπ₯ by dπ¦ is equal to π to the power of π¦.
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Now, weβre trying to find the differential of the reciprocal function, so thatβs π¦, with respect to π₯.
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So thatβs dπ¦ by dπ₯.
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And in order to do this, we can use the fact that the derivative of an inverse of a function is equal to the reciprocal of the derivative of the function.
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Giving us that dπ¦ by dπ₯ is equal to one over dπ₯ by dπ¦.
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In order to use this, we must ensure that the denominator of our fraction is nonzero.
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So thatβs dπ₯ by dπ¦.
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Weβve just found that dπ₯ by dπ¦ is equal to π to the power of π¦.
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Since π to the π¦ is an exponential, we know that π to the power of π¦ is going to be greater than zero for all values of π¦.
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Therefore, it is nonzero.
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And so, weβre able to use this formula.
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And so, we obtain that dπ¦ by dπ₯ is equal to one over π to the power of π¦.
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However, the question has asked us to give our answer in terms of π₯.
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In order to get our answer in terms of π₯, we can use the fact that π₯ is equal to π to the power of π¦ and substitute π₯ in for π to the power of π¦.
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From here, we reach our solution, which is that dπ¦ by dπ₯ is equal to one over π₯.
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Letβs quickly stop to think about what weβre shown here.
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Our original function is π₯ is equal to π to the power of π¦.
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We can make π¦ the subject of this equation.
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We simply take natural logs of both sides.
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This gives us that π¦ is equal to the natural logarithm of π₯.
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And this is the inverse of the function given in the question.
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Now, we have found dπ¦ by dπ₯.
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Since π¦ is equal to the natural logarithm of π₯, dπ¦ by dπ₯ is the differential of the natural logarithm of π₯ with respect to π₯.
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Therefore, we have just shown that differential of the natural logarithm of π₯ with respect to π₯ is equal to one over π₯.
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We will now consider another example.
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Given that π₯ is equal to π¦ to the power of five plus the square root of π¦ plus the cube root of π¦ squared, find dπ¦ by dπ₯.
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We can find dπ¦ by dπ₯ by using the formula for the derivative of the inverse function.
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Which is that dπ¦ by dπ₯ is equal to one over dπ₯ by dπ¦.
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So we start by differentiating π₯ with respect to π¦.
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Letβs start by rewriting some of the terms in π₯.
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We can rewrite the square root of π¦ as π¦ to the power of a half and the cube root of π¦ squared as π¦ to the power of two over three.
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Now, we can use the power rule for differentiation in order to differentiate π₯ with respect to π¦, term by term.
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We multiply by the power and decrease the power by one.
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This gives us that dπ₯ by dπ¦ is equal to five π¦ to the power of four plus one-half π¦ to the power of negative one-half plus two-thirds π¦ to the power of negative one-third.
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We can rewrite that fractional powers of π¦ back in their surd form.
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And then, we can combine these three terms into one fraction by creating a common denominator of six π¦.
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Our first term becomes 30π¦ to the power of five over six π¦.
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Our second term becomes three multiplied by the square root of π¦ over six π¦.
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And our third term becomes four multiplied by the cube root of π¦ squared over six π¦.
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We obtain that dπ₯ by dπ¦ is equal to 30 multiplied by π¦ to the power of five plus three multiplied by the square root of five plus four multiplied by the cube root of π¦ squared all over six π¦.
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Now, we can apply the formula for the derivative of the inverse function.
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And this gives us our solution that dπ¦ by dπ₯ is simply the reciprocal of dπ₯ by dπ¦.
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Sometimes, we may be asked to find the derivative of the inverse function at a given point.
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We must be careful about which point we evaluate the function, as the next example will show.
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Let π of π₯ be equal to one-half π₯ cubed plus one-half π₯ squared plus five π₯ minus four and let π be the inverse of π.
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Given that π of two is equal to 12, what is π prime of 12?
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In order to help us find π prime of 12, we can use the formula for derivatives of inverse functions.
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This tells us that if π is the inverse function of π, then π prime of π¦ is equal to one over π prime of π of π¦.
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Letβs start by finding π prime of π₯, the derivative of π with respect to π₯.
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We can see that π is a polynomial.
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Therefore, in order to find its derivative, we can differentiate it term by term using the power rule for differentiation.
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We simply multiply by the power and decrease the power by one.
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This gives us that π prime of π₯ is equal to three over two π₯ squared plus π₯ plus five.
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Next, weβll observe the fact that weβre trying to find π prime of 12.
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And so, we can substitute π¦ equals 12 into our formula for π prime of π¦.
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This gives us that π prime of 12 is equal to one over π prime of π of 12.
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Now, we do not know what π of 12 is.
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However, we have been given in the question that π of two is equal to 12.
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And since π is the inverse function of π, we can apply π to both sides here.
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And weβll obtain that π of 12 is equal to two.
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This is because of the way inverse functions work.
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If we take π of π of two, then weβll simply get two.
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We can now substitute this value of π of 12 back into our equation for π prime of 12.
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And we obtain that itβs equal to one over π prime of two.
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Now, weβve already found π prime of π₯.
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So we can simply substitute π₯ equals two in order to find π prime of two.
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And we obtain that π prime of two is equal to 13.
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And substituting the value of π prime of two back into π prime of 12, we obtain that π prime of 12 is equal to one over 13.
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Now, our original definition for the derivative of an inverse function assumed that the inverse existed.
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We will now cover what is called the inverse function theorem, which is more powerful than our original definition.
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Since it guarantees the existence and continuity of the inverse of a function when it is continuously differentiable with a nonzero derivative.
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The Inverse Function Theorem
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Let π be a continuously differentiable function with a nonzero derivative at a point π.
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Then the inverse function theorem tells us that: One, π is invertible in a neighborhood of π.
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Two, π has a continuously differentiable inverse in a neighborhood of π.
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Three, the derivative of the inverse of the point π is equal to π of π is equal to the reciprocal of the derivative of π at π.
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That is that the derivative of the inverse of π of π is equal to one over the derivative of π of π.
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All we require to use this theorem is for π to be continuously differentiable and to have a nonzero derivative at some point π.
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Now, the proof of this theorem is beyond the scope of this video.
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So we will not be covering it here.
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We will now move on and look at some more examples.
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If π of two π is equal to negative one, π prime of two π is equal to one, and π is equal to negative one, find the derivative of the inverse of π at π.
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Weβll be using the fact that the derivative of the inverse of π at π is equal to one over π prime of π inverse of π.
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In our case, π is equal to negative one.
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We need to start by finding π inverse of negative one.
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Weβre given in the question that π of two π is equal to negative one.
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Since we know that π inverse is the inverse function of π, this tells us that π inverse of negative one is equal to two π.
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So we can substitute this into our equation.
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And now, we have that the derivative of π inverse at negative one is equal to one over π prime of two π.
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And we can see that weβve actually been given π prime of two π in the question.
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And itβs equal to one.
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So we can substitute this in.
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And weβve reached our solution, which is that the derivative of the inverse function of π at negative one is equal to one.
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We will now look at one final example.
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Let π be the inverse of π.
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Using the table below, find π prime of zero.
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In order to find π prime of zero, weβll be using the formula for finding the derivative of an inverse of a function.
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Which tells us that π prime of π¦ is equal to one over π prime of π of π¦.
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Weβre trying to find π prime of zero.
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So we can substitute in zero for π¦.
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This gives us that π prime of zero is equal to one over π prime of π of zero.
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From the table, we can see that when π₯ is equal to zero, π is equal to negative one.
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And so, we have that π of zero is equal to negative one.
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Which we can substitute in to give us that π prime of zero is equal to one over π prime of negative one.
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And now, we can simply read π prime of negative one off from the table.
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Since when π₯ is equal to negative one, π prime is equal to one-third.
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Giving us that π prime of negative one is equal to one-third.
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And again, this can be substituted in.
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And so, we will obtain that π prime of zero is equal to the reciprocal of one-third.
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And here, we reach our solution of three.
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We have now learned about derivatives of inverse functions.
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And weβve seen a variety of examples of how they work.
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Letβs recap some key points of this video.
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Key Points
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Given a continuously differentiable function π with a nonzero derivative at a point π.
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The derivative of the inverse of the function at π, which is equal to π of π, is the derivative of the inverse of π at π is equal to one over the derivative of π at π.
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This is often written in Leibnizβs notation as dπ¦ by dπ₯ is equal to one over dπ₯ by dπ¦.
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We need to be careful about which points weβre using.
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When using these equations, we can find derivatives of many familiar inverse functions, such as the natural logarithm.
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The inverse function theorem guarantees the existence of the inverse of a continuous function around points with nonzero derivatives.
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Using this theorem, we can find the derivatives of inverse functions.
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Even when we are unable to find an explicit formula for the inverse.