“Teach A Level Maths” Vol. 2: A2 Core Modules 38: Implicit Differentiation © Christine Crisp

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We usually write a relationship between x and y in the form This gives y explicitly in terms of x. However, we’ve already met some curves, for example, a circle, where it is easier to have x and y “mixed up” on the same side of the equation. These give y implicitly. e.g. is explicit is implicit

It’s not always easy to rearrange implicit formulae to explicit form so we need to be able to differentiate them as they are. We’ll first think about differentiating ( explicit ) We get This is called differentiating with respect to (w.r.t.) x So, differentiating y w.r.t. x gives

It’s not always easy to rearrange implicit formulae to explicit form so we need to be able to differentiate them as they are. We’ll first think about differentiating ( explicit ) We get This is called differentiating with respect to (w.r.t.) x So, differentiating y w.r.t. x gives Also, differentiating with respect to x is easy.

It isn’t usually easy to rearrange implicit formulae to explicit form so we need to be able to differentiate them as they are. We’ll first think about differentiating ( explicit ) We get This is called differentiating with respect to (w.r.t.) x So, differentiating y w.r.t. x gives Also, differentiating with respect to x is easy. So, differentiating with respect to y is also easy. It gives 2y.

Differentiating with respect to y gives 2y. To differentiate with respect to x we use the chain rule. The chain rule for differentiating : becomes We choose y as the “chaining variable” here . . .

Differentiating with respect to y gives 2y. To differentiate with respect to x we use the chain rule. The chain rule for differentiating : becomes We choose y as the “chaining variable” here . . . because we’ve just seen that So, or

y w.r.t. x, giving We can now differentiate the following: and as before: w.r.t. x, giving y w.r.t. x, giving and Tip: Whenever we have a function of y to differentiate w.r.t. x, we just differentiate it, using the usual rules, and multiply by .

e.g. 1 Differentiate the following with respect to x (b) Solution: (a) (b)

e.g. 2 A curve is defined by the equation Find the gradient at the point . Solution: Tip: Don’t rearrange to find unless you are asked to. Just substitute.

This means the function on the r.h.s. can contain x and y. e.g. 3 Find the gradient function of the curve giving your answer in the form This means the function on the r.h.s. can contain x and y. Solution:

Exercise 1. Differentiate the following with respect to x: (a) (b) 2. Find the gradient at ( -1, -3 ) on the curve

1(a) Solution: (b) 2. Find the gradient at ( -1, -3 ) on the curve Solution:

We may have to differentiate terms such as xy. This is a product so we use the product rule: becomes Tip: With implicit equations I always look for a product and if I see one I write P by it so that I don’t then forget it !

Can you see how we could make a mistake with this product? e.g. 1 Given show that P Solution: Can you see how we could make a mistake with this product? ANS: The minus sign. There will be 2 terms so we must use brackets. So,

In this question, we do have to rearrange to find Collect terms containing on the l.h.s. and the others on the r.h.s. Common factor: Divide by ( 2y - x ):

e.g. 2 The equation of a curve is Find an equation connecting x, y and at all points on the curve. Hence show that the coordinates of the points on the curve at which satisfy the equation Solution: P

e.g. 3 Find an equation linking x, y and if P Solution:

Exercise 2. Find the gradients at the 2 points on the curve given by where . 1. A curve is given by the equation ( You will need to find the values of y at x = 1. ) Find an equation linking x, y and and find the value of at the point ( 0, 2 ).

Solutions: 1. P The gradient at (0, 2) is 1.

P 2. The gradients are 6 and -1

Exercise 3. A curve is given by the equation Find an equation linking x, y and . Solution: P

One useful application of implicit differentiation may arise in growth and decay problems which we look at later. e.g. Suppose we want to find given that We can’t differentiate when x is an index ( apart from ), so we have to take logs. Using base e: Using the 3rd law of logs: Differentiate w.r.t. x: is just a constant Substitute for y:

Generalising the last result:

The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied. For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.

We usually write a relationship between x and y in the form However, we’ve already met some curves, for example, a circle, where it is easier to have x and y “mixed up” on the same side of the equation. This gives y explicitly in terms of x. e.g. is explicit is implicit These give y implicitly.

To differentiate with respect to x we use the chain rule. The chain rule for differentiating : becomes We choose y as the “chaining variable” here . . . Differentiating with respect to y gives 2y. So, or because we’ve just seen that

y w.r.t. x, giving We can now differentiate the following: and as before: y w.r.t. x, giving and Tip: Whenever we have a function of y to differentiate w.r.t. x, we just differentiate it, using the usual rules, and multiply by .

Solution: e.g. A curve is defined by the equation Find the gradient at the point . Tip: Don’t rearrange to find unless you are asked to. Just substitute.

We may have to differentiate terms such as xy. This is a product so we use the product rule: becomes Tip: With implicit equations I always look for a product and if I see one I write P by it so that I don’t then forget it !

e.g. The equation of a curve is Solution: Find an equation connecting x, y and at all points on the curve. Hence show that the coordinates of the points on the curve at which satisfy the equation P