Video: Implicit Differentiation

In this video, we will learn how to use implicit differentiation to differentiate functions defined implicitly.

17:02

Video Transcript

In this video, we’ll learn how to use implicit differentiation to help us find the derivative of functions expressed implicitly as functions of π‘₯. The majority of differentiation problems you will have come across so far will have involved functions written explicitly as functions of π‘₯, such as 𝑦 equals three π‘₯ squared sin π‘₯. In this video, we’re going to explore how implicit differentiation, which is an extension of the chain rule, allows us to easily differentiate equations such as the Cartesian equation of a circle, π‘₯ squared plus 𝑦 squared equals one for example. We’ll also explore what this means for the second derivative and higher order derivatives.

The chain rule allows us to differentiate composite functions. It says that for two differentiable functions, 𝑔 and β„Ž, such that 𝑓 is the composite function 𝑔 of β„Ž of π‘₯, the derivative of 𝑓 is the derivative of β„Ž of π‘₯ times the derivative of 𝑔 evaluated at β„Ž of π‘₯. It’s often more intuitive though to write this as d𝑦 by dπ‘₯ equals d𝑒 by dπ‘₯ times d𝑦 by d𝑒, where 𝑦 is a function of 𝑒 and 𝑒 is a function of π‘₯. Let’s demonstrate how this chain rule can help us to find the derivative of an implicit function.

Consider the equation π‘₯ squared plus 𝑦 squared equals one. Using implicit differentiation, find an expression for d𝑦 by dπ‘₯ in terms of π‘₯ and 𝑦. And there’s a second part to this question. For the semicircle where 𝑦 is greater than or equal to zero, express 𝑦 explicitly in terms of π‘₯, then differentiate this expression to get an expression for d𝑦 by dπ‘₯ in terms of π‘₯.

We’ll begin with part one. To differentiate the function implicitly, we’re going to begin by differentiating both sides of the equation with respect to π‘₯. This is probably going to look a little strange, but bear with me. We say that the derivative of π‘₯ squared plus 𝑦 squared with respect to π‘₯ is equal to the derivative of one with respect to π‘₯. And we absolutely must do this to both sides of the equation. Next, we differentiate what we can. It’s quite straightforward to differentiate one with respect to π‘₯. It’s simply zero. But how do we differentiate π‘₯ squared plus 𝑦 squared?

Well, the derivative of π‘₯ squared is two π‘₯. But the derivative of 𝑦 squared is a little more strange. We know that 𝑦 squared is a function of 𝑦. And, in turn, 𝑦 is a function of π‘₯. So here, we can use the chain rule. We say that the derivative of 𝑦 squared with respect to π‘₯ is equal to the derivative of 𝑦 squared with respect to 𝑦 times the derivative of 𝑦 with respect to π‘₯. Well, the derivative of 𝑦 squared with respect to 𝑦 is two 𝑦. And the derivative of 𝑦 with respect to π‘₯ is simply d𝑦 by dπ‘₯. So our equation now becomes two π‘₯ plus two 𝑦 d𝑦 by dπ‘₯ equals zero.

Remember, we want an equation for the derivative. So we’re going to make d𝑦 by dπ‘₯ the subject by first subtracting two π‘₯ from both sides of our equation. That gives us two 𝑦 d𝑦 by dπ‘₯ is equal to negative two π‘₯. Next, we’re going to divide through by two 𝑦. And we see that d𝑦 by dπ‘₯ is equal to negative two π‘₯ divided by two 𝑦. Well, the twos cancel. And we found an expression for d𝑦 by dπ‘₯ in terms of π‘₯ and 𝑦; it’s negative π‘₯ over 𝑦.

For part two of this question, we’re going to go back to our original equation. And we’re going to begin by making 𝑦 the subject of the equation. That’s expressing 𝑦 explicitly in terms of π‘₯. We can subtract π‘₯ squared from both sides. And then we take the square root of both sides of the equation. Now, usually, we take by the positive and negative square root. Here though, we’re told the semicircle is such that 𝑦 is greater than or equal to zero. So we’re just going to consider the positive square root. And we have an explicit function in π‘₯.

Now, writing it as one minus π‘₯ squared to the power of a half, we can see that we can use the general power rule to differentiate this. This says that if 𝑒 is some function in π‘₯, the derivative of 𝑒 to the power of 𝑛 with respect to π‘₯ is equal to 𝑛 times 𝑒 to the power of 𝑛 minus one multiplied by the derivative of 𝑒 with respect to π‘₯. And this is of course when 𝑛 is a real number. So the derivative of one minus π‘₯ squared to the power of a half is a half times one minus π‘₯ squared to the power of negative half multiplied by d𝑒 by dπ‘₯. But, actually, 𝑒 is equal to one minus π‘₯ squared. So the derivative of 𝑒 with respect to π‘₯ is just negative two π‘₯. Once again, we have some twos to cancel. And our expression for d𝑦 by dπ‘₯ in terms of π‘₯ is negative π‘₯ over the square root of one minus π‘₯ squared.

Notice that since we said that 𝑦 was equal to the square root of one minus π‘₯ squared, we could have written this as d𝑦 by dπ‘₯ equals negative π‘₯ over 𝑦. And that’s the same answer as we got in part one of this question. And, of course, we must remember that 𝑦 cannot be equal to zero here. Now, this example demonstrates some important points. Firstly, even though it was easy enough to express this relation as an explicit function, it was simpler to differentiate using implicit differentiation.

And, in general, when we’re differentiating implicitly, we can use the following version of the chain rule. This says the derivative of a function in 𝑦 with respect to π‘₯ is equal to the derivative of that function in 𝑦 with respect to 𝑦 times d𝑦 by dπ‘₯. It’s really useful to commit this version of the chain rule to memory. Let’s now apply it to a more complicated example.

Find d𝑦 by dπ‘₯ by implicit differentiation if negative 𝑒 of to the 𝑦 sin π‘₯ is equal to four π‘₯𝑦 plus two π‘₯.

To differentiate this function implicitly, we begin by differentiating both sides of the equation with respect to π‘₯. We begin by writing this as d by dπ‘₯ of negative 𝑒 to the 𝑦 times sin π‘₯ equals d by dπ‘₯ of four π‘₯𝑦 plus two π‘₯. Let’s differentiate each bit with respect to π‘₯. Well, the derivative of two π‘₯ is straightforward; it’s simply two. We’re going to have to use the product rule combined with the chain rule to differentiate four π‘₯𝑦 and negative 𝑒 to the 𝑦 sin π‘₯. The special version of the chain rule that we need says the derivative of 𝑓 of 𝑦 with respect to π‘₯ is equal to the derivative of 𝑦 with respect to 𝑦 times d𝑦 by dπ‘₯. And the product rule says that the derivative of 𝑒 times 𝑣 is equal to 𝑒 times the derivative of 𝑣 plus 𝑣 times the derivative of 𝑒.

We’ll begin by differentiating four π‘₯𝑦. We’re going to let 𝑒 be equal to four π‘₯ and 𝑣 be equal to 𝑦. Then the derivative of 𝑒 with respect to π‘₯ is simply four. The derivative of 𝑣 with respect to π‘₯ is equal to the derivative of 𝑦 with respect to 𝑦 which is one times d𝑦 by dπ‘₯ which is simply d𝑦 by dπ‘₯. 𝑒 times d𝑣 by dπ‘₯ is then four π‘₯ multiplied by d𝑦 by dπ‘₯. And 𝑣 multiplied by d𝑒 by dπ‘₯ is equal to 𝑦 times four. And, therefore, the right-hand side of our equation is four π‘₯ d𝑦 by dπ‘₯ plus four 𝑦 plus two. Let’s now repeat this process for negative 𝑒 to the power of 𝑦 times sin π‘₯.

This time, let’s say 𝑒 is equal to negative 𝑒 to the power of 𝑦 and 𝑣 is equal to sin π‘₯. The derivative of sin π‘₯ is cos π‘₯. And then the derivative of 𝑒 with respect to π‘₯ is equal to the derivative of negative 𝑒 to the 𝑦 with respect to 𝑦, that’s negative 𝑒 to the 𝑦, times d𝑦 by dπ‘₯. And then when we use the product rule, we see that d𝑦 by dπ‘₯ is equal to negative 𝑒 to the 𝑦 cos π‘₯ minus 𝑒 to the 𝑦 sin π‘₯ d𝑦 by dπ‘₯. So our equation is now as shown. Remember, we’re trying to find an equation for d𝑦 by dπ‘₯. So we’re going to rearrange and make d𝑦 by dπ‘₯ the subject. When we do, we see that negative 𝑒 to the 𝑦 cos π‘₯ minus four 𝑦 minus two is equal to four π‘₯ plus 𝑒 to the 𝑦 sin π‘₯ all multiplied by d𝑦 by dπ‘₯.

Now, we can actually factor negative one on the left-hand side. And then, we divide through by four π‘₯ plus 𝑒 to the 𝑦 sin π‘₯. And we see that d𝑦 by dπ‘₯ is equal to negative 𝑒 to the 𝑦 cos π‘₯ plus four 𝑦 plus two over four π‘₯ plus 𝑒 to the 𝑦 sin π‘₯. And, of course, this derivative stands when the denominator is not equal to zero, when four π‘₯ plus 𝑒 to the 𝑦 sin π‘₯ is not equal to zero. It’s fairly common to use implicit differentiation to find the equation of a tangent to a curve defined implicitly. In our next example, we’ll look at how we can use implicit differentiation to address a problem of this type.

The equation 𝑦 squared minus 24π‘₯ cubed plus 24π‘₯ equals zero describes a curve in the plane. 1) Find the coordinates of two points on this curve, where π‘₯ equals negative one-half. 2) Determine the equation of the tangent at the points where π‘₯ equals negative one-half and the 𝑦-coordinate is positive. 3) Find the coordinates of another point, if it exists, at which the tangent meets the curve.

For part one, to find the points where π‘₯ is equal to negative one-half, we’re going to substitute this value of π‘₯ into our equation and solve for 𝑦. That’s 𝑦 squared minus 24 times negative a half cubed plus 24 times negative a half. And that’s equal to zero. When we evaluate this, we get 𝑦 squared plus three minus 12 equals zero or 𝑦 squared minus nine equals zero. We’re going to solve this equation by adding nine to both sides.

And our last step is to find the square root of both sides of this equation, remembering to take by the positive and negative square roots of nine. The square root of nine is three. So we see that 𝑦 is equal to three and negative three, when π‘₯ is equal to negative one-half. In coordinate form, that’s negative a half, three and negative a half, negative three.

We now consider part two. We need to find the equation of the tangent at a point where π‘₯ equals negative a half and the 𝑦-coordinate is positive. That’s the coordinate negative a half, three. We’ll first, however, need to find the gradient of the tangent to the curve. This will be the derivative of the equation for the curve evaluated at π‘₯ equals negative a half and 𝑦 equals three. So we’re going to differentiate our equation implicitly. That’s d by dπ‘₯ of 𝑦 squared minus 24π‘₯ cubed plus 24π‘₯ equals d by dπ‘₯ of zero. The derivative of 𝑦 squared is the derivative of 𝑦 squared with respect to 𝑦 times the derivative of 𝑦 with respect to π‘₯. It’s two 𝑦 d𝑦 by dπ‘₯. The derivative of negative 24π‘₯ cubed is three times negative 24π‘₯ squared. That’s negative 72π‘₯ squared. The derivative of 24π‘₯ is 24. And the derivative of zero is zero.

So we see that two 𝑦 d𝑦 by dπ‘₯ minus 72π‘₯ squared plus 24 is equal to zero. We need an equation for d𝑦 by dπ‘₯. So we’re going to rearrange to make d𝑦 by dπ‘₯ the subject. We add 72π‘₯ squared to both sides of this equation and subtract 24. Next, we’re going to divide through by two 𝑦. And we see that d𝑦 by dπ‘₯ is equal to 72π‘₯ squared minus 24 over two 𝑦, which simplifies to 36π‘₯ squared minus 12 over 𝑦. Remember, we want to find the gradient of the tangent of the curve at negative a half, three. So we’re going to substitute π‘₯ equals negative a half and 𝑦 equals three into the equation for the derivative. When we do, we see that the gradient of our tangent is 36 times negative a half squared minus 12 all over three, which is negative one.

Finally, we substitute what we know about our tangent into the equation of a straight line. And we get 𝑦 minus three equals negative one times π‘₯ minus negative a half. Distributing the parentheses and simplifying then rearranging for 𝑦, we get 𝑦 equals five over two minus π‘₯.

And now we consider part three. We need to find a point, if it exists, where this tangent meets the curve again. We therefore want to solve simultaneously the equations 𝑦 squared minus 24π‘₯ cubed plus 24π‘₯ equals zero and 𝑦 equals five over two minus π‘₯. We can achieve this by substituting 𝑦 equals five over two minus π‘₯ into the original equation for the curve. That gives us five over two minus π‘₯ all squared minus 24π‘₯ cubed plus 24π‘₯ equals zero. Expanding the parentheses and multiplying through by negative one, and we end up with the equation shown.

Next, we could solve this using a scientific calculator. Alternatively, we know that π‘₯ equals negative a half is a root to this equation. And we can therefore factor out π‘₯ plus a half. And we can do this by using long division or matching coefficients. The first step towards equating coefficients would be to write the equation shown. And whilst it’s outside the scope of this video to spend a lot of time performing this, we should find that π‘Ž is equal to 24, 𝑏 is negative 13, and 𝑐 is negative 25 over two. And you can pause the video here if you like and see if you can complete that step yourself.

Our last step is to solve the quadratic equation 24π‘₯ squared minus 13π‘₯ minus 25 over two. And we could do that by using the quadratic formula or completing the square. When we solve this, we see that π‘₯ is equal to negative a half. So that’s a repeated root. And π‘₯ is equal to 25 over 24. We’re finding the coordinates, so we substitute π‘₯ equals 25 over 24 into the equation five over two minus π‘₯. And that gives us a 𝑦-value of 35 over 24. And so the tangent with equation 𝑦 equals five over two minus π‘₯ meets the curve at negative a half, three and twenty-five twenty-fourths, thirty-five twenty-fourths. In our very final example, we’ll look at how implicit differentiation can help us to find higher order derivatives.

Given that two sin 𝑦 minus five cos π‘₯ equals negative four, determine the second derivative of 𝑦 by implicit differentiation.

To find to the second derivative of 𝑦, sometimes called 𝑦 double prime, we’re going to need to differentiate our function twice. Notice how the function itself is expressed implicitly as functions of π‘₯. So we’re going to use implicit differentiation and begin by finding the derivative of both sides. Now the derivative of negative four with respect to π‘₯ is fairly easy to work out. It’s just zero. Similarly, we can find the derivative of negative five cos π‘₯ with respect to π‘₯. That’s five sin π‘₯. But what about the derivative of two sin 𝑦 with respect to π‘₯? Well, here we use a special version of the chain rule. This says that the derivative of a function in 𝑦 with respect to π‘₯ is equal to the derivative of that function with respect to 𝑦 times d𝑦 by dπ‘₯.

Well, the derivative of two sin 𝑦 with respect to 𝑦 is two cos 𝑦. So our equation is two cos 𝑦 d𝑦 by dπ‘₯ plus five sin π‘₯. And that’s equal to zero. We can find the first derivative then by subtracting five sin π‘₯ from both sides of the equation and dividing through by two cos 𝑦. Remember, though, we were looking to find the second derivative. So here we’re going to need to use the quotient rule to differentiate negative five sin π‘₯ over two cos 𝑦. According to our notation, we can let 𝑒 be equal to negative five sin π‘₯ and 𝑣 equal two cos 𝑦. d𝑒 by dπ‘₯ is negative five cos π‘₯. Then since the derivative of 𝑣 with respect to 𝑦 is negative two sin 𝑦, we can see that d𝑣 by dπ‘₯ is negative two sin 𝑦 d𝑦 by dπ‘₯.

We substitute each of these into the formula for the quotient rule. We then simplify and spot that we said that d𝑦 by dπ‘₯ is negative five sin π‘₯ over two cos 𝑦. So we can substitute this into the formula for the second derivative. To simplify this somewhat, we multiply both the numerator and denominator of our fraction by two cos 𝑦. And we’re going to look to simplify just a little more. Distributing this negative one, we can see that the second derivative of our function is 25 sin squared π‘₯ sin 𝑦 minus 10 cos π‘₯ cos squared 𝑦 all over cos cubed 𝑦. And this formula is valid as long as cos 𝑦 is not equal to zero.

In this video, we’ve seen that when given a function defined implicitly, we can use a special version of the chain rule to differentiate it. The derivative of a function in 𝑦 with respect to π‘₯ is equal to the derivative of that function 𝑦 with respect to 𝑦 times d𝑦 by dπ‘₯. We also saw that if it was possible to rewrite a relation as an explicit function, it was sometimes simpler to use implicit differentiation. We also saw that when we differentiate implicitly, we’ll get an expression for d𝑦 by dπ‘₯ in terms of both π‘₯ and 𝑦. We also saw that we can find higher order derivatives by using implicit differentiation. And in these cases, we’ll need to substitute expressions for lower order derivatives to help us simplify the expressions.

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