Lesson Explainer: Implicit Differentiation | Nagwa Lesson Explainer: Implicit Differentiation | Nagwa

Lesson Explainer: Implicit Differentiation Mathematics

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

By this point in a calculus course, we are able to differentiate many types of functions where one variable is explicitly expressed in terms of another. For example, 𝑦=1𝑥,𝑦=𝑒𝑥.cos

Generally, they are functions of the type 𝑦=𝑓(𝑥). However, some functions are not defined explicitly like this but are instead defined implicitly by a relation between 𝑥 and 𝑦. One familiar example would be the equation of a circle 𝑥+𝑦=𝑟, which is implicitly defining a function. In this case, it is possible to solve for 𝑦 to get two explicit functions of 𝑥; namely, 𝑦=𝑟𝑥,𝑦=𝑟𝑥.

However, this might not be possible in general. For example, consider the relation 𝑦𝑦=𝑥𝑥.sinsin

It is not possible to find a simple closed-form expression explicitly defining 𝑦 in terms of 𝑥. However, by using implicit differentiation, we are still able to find an expression for the derivative of 𝑦 with respect to 𝑥 even when we cannot rearrange a relation to get an explicit function. How will we be able to do this? We will need to apply the chain rule.

Rule: Chain Rule

Given a function 𝑓(𝑦) of variable 𝑦, its derivative of with respect to the variable 𝑥 is given by dddd𝑥𝑓(𝑦)=𝑓(𝑦)𝑦𝑥.

Here 𝑓(𝑦) is the derivative of the function 𝑓(𝑦) obtained using the rules of differentiation.

When the variable of the function we are differentiating is 𝑥, then the derivative dd𝑥 can be immediately computed using the rules of differentiation. However, if the variable of the function we are differentiating is 𝑦, then the chain rule stated above tells us that the derivative involves an extra terms, dd𝑦𝑥.

We will demonstrate the method by example.

Example 1: Implicit Differentiation

Consider the equation 𝑥+𝑦=1.

  1. Using implicit differentiation, find an expression for dd𝑦𝑥 in terms of 𝑥 and 𝑦.
  2. For the semicircle where 𝑦0, express 𝑦 explicitly in terms of 𝑥; then, differentiate this expression to get an expression for dd𝑦𝑥 in terms of 𝑥.

Answer

Part 1

To implicitly differentiate, we need to differentiate both sides of the equation with respect to 𝑥 as follows: dddd𝑥𝑥+𝑦=𝑥(1).

Using the linearity of the derivative, we can rewrite this as dddd𝑥𝑥+𝑥𝑦=0.

Using the power rule, we can differentiate the first term:

2𝑥+𝑥𝑦=0.dd(1)

As for the second, since 𝑦 is a function of 𝑦 and we can consider 𝑦 to be a function of 𝑥, we can use the chain rule to get dddd𝑥𝑦=2𝑦𝑦𝑥.

Substituting this back into equation (1), we have 2𝑥+2𝑦𝑦𝑥=0.dd

We can now rearrange to make dd𝑦𝑥 the subject. By subtracting 2𝑥 from both sides, and dividing by two, we have 𝑦𝑦𝑥=𝑥.dd

For all 𝑦0, we can rewrite this as dd𝑦𝑥=𝑥𝑦.

Part 2

We begin by rearranging the equation we have been given to make 𝑦 the subject. By subtracting 𝑥 from both sides of the equation, we have 𝑦=1𝑥.

Since we are only interested in the semicircle where 𝑦0, we can take the positive square root of both sides of the equation to get 𝑦=1𝑥.

Now we have an explicit formula for 𝑦 in terms of 𝑥, we can differentiate using the familiar rules of derivatives. Notice that we have a composition of functions, so we will need to apply the chain rule. Setting 𝑢=1𝑥, we have 𝑦=𝑢. We now differentiate to get dddd𝑦𝑢=12𝑢,𝑢𝑥=2𝑥.

Applying the chain rule, dddddd𝑦𝑥=𝑦𝑢𝑢𝑥, we have dd𝑦𝑥=121𝑥(2𝑥)=𝑥1𝑥.

Notice that since 𝑦=1𝑥, we could express this as dd𝑦𝑥=𝑥𝑦, which is the same answer we got in the first part of the question.

The last example demonstrates a number of important points:

  • Even when it is possible to rewrite a relation as an explicit function, it is often simpler to use implicit differentiation.
  • When we differentiate implicitly, in general, we will get an expression for dd𝑦𝑥 in terms of both 𝑥 and 𝑦.

It is fairly common to use implicit differentiation to find the equation of a tangent to a curve defined implicitly. In the next couple of examples, we will demonstrate how we can use implicit differentiation to address problems of this type.

Example 2: Finding Tangents to Implicitly Defined Curves

The equation 𝑦24𝑥+24𝑥=0 describes a curve in the plane.

  1. Find the coordinates of two points on this curve, where 𝑥=12.
  2. Determine the equation of the tangent at the points where 𝑥=12 and the 𝑦-coordinate is positive.
  3. Find the coordinates of another point, if it exists, at which the tangent meets the curve.

Answer

Part 1

To find the points on the curve where 𝑥=12, we substitute this value into the equation and solve for 𝑦. Hence, 0=𝑦2412+2412=𝑦+312=𝑦9.

Hence, 𝑦=9, which implies 𝑦=3 and 𝑦=3 are the two 𝑦-values where 𝑥=12, which we can write as coordinates as 12,3 and 12,3.

Part 2

We need to find the equation of the tangent to the curve at 12,3. To do this, we start by finding the derivative at this point because this will be equal to the slope of the tangent. Hence, differentiating implicitly, we have 2𝑦𝑦𝑥72𝑥+24=0.dd

Hence, 2𝑦𝑦𝑥=72𝑥24.dd

Since, for the point we are interested in, 𝑦0, we can rewrite this as dd𝑦𝑥=36𝑥12𝑦.

Substituting in 𝑥=12 and 𝑦=3, we have dd𝑦𝑥|||=36123=1244=1.()()

Therefore, the slope of the tangent at 12,3 is 1. Hence, using the slope–point form of the equation of a line, 𝑦𝑦=𝑚(𝑥𝑥), we can find the equation of the tangent as follows: 𝑦3=1𝑥12.

We can simplify this to 𝑦=52𝑥.

Part 3

To find the coordinates of another point where this tangent crosses the curve, we can substitute 𝑦=52𝑥 into the equation for the curve and try to solve for 𝑥. We can then find the corresponding value of 𝑦. Therefore, 52𝑥24𝑥+24𝑥=0.

Expanding the parentheses, we have 2545𝑥+𝑥24𝑥+24𝑥=0.

Hence, 24𝑥𝑥19𝑥254=0.

We already know that the tangent passes through the point 12,3. Hence, 𝑥=12 is one of the roots of this equation. We can, therefore, factor this out. We can accomplish this via long division or by matching the coefficients of the terms. We start by writing 𝑥+12𝑎𝑥+𝑏𝑥+𝑐=24𝑥𝑥19𝑥254.

We can see that the first term in the parentheses must be 24𝑥 to ensure we get the correct coefficient for the 𝑥-term. Hence, 𝑥+1224𝑥+𝑏𝑥+𝑐=24𝑥𝑥19𝑥254.

Similarly, we can see that the constant term in the parentheses must be 252. Therefore, 𝑥+1224𝑥+𝑏𝑥252=24𝑥𝑥19𝑥254.

Finally, we need to find the middle term in the parentheses. We can do this by comparing the coefficients of the 𝑥-term on both sides of the equation. On the right-hand side, we have a coefficient of 1, whereas on the left-hand side, we have a coefficient of 𝑏+12. Therefore, 𝑏=13 and 𝑥+1224𝑥13𝑥252=24𝑥𝑥19𝑥254.

It is often useful to check that this value gives us the correct coefficient for the final term on the right-hand side, which is 19𝑥. Therefore, considering the coefficient of 𝑥 on the left-hand side, we have 132252=19 as required.

We now have a quadratic equation we can solve. Using the quadratic formula or otherwise, we find this has solutions 𝑥=12 and 𝑥=2524. Therefore, the tangent intersects the curve again at the point where 𝑥=2524. Using the equation of the tangent, we can substitute in this value of 𝑥 to find the corresponding value of 𝑦 as follows: 𝑦=522524=3524.

Hence, the tangent at 12,3 meets the curve again at the point 2524,3524.

Sometimes, we may be asked to find the equation of a tangent from its slope as opposed to the point of tangency. In the next example, we will tackle a problem of this type.

Example 3: Finding Tangents with Given Slopes

Find, for 0<𝑥<𝜋, the tangent to 9𝑦=(5𝑥+3𝑦)cos that has slope 512, giving your equation in terms of 𝜋.

Answer

We have been given the slope of a tangent and would like to find the values of 𝑥 and 𝑦 at the point that has this slope. To do this, we begin by differentiating the equation to find an expression for the derivative dd𝑦𝑥. Using implicit differentiation, 9𝑦𝑥=5+3𝑦𝑥(5𝑥+3𝑦).ddddsin

Since the derivative is equal to the slope of the tangent, we can substitute 512 in for dd𝑦𝑥 as follows: 9512=5+3512(5𝑥+3𝑦).sin

Simplifying, we get 154=154(5𝑥+3𝑦)1=(5𝑥+3𝑦).sinsin

Hence, 5𝑥+3𝑦=𝜋2.

In previous examples, we computed the first derivative dd𝑦𝑥 and applied the derivative to find the slope of the tangent line at a point. Sometimes, we are interested in finding the second, or higher-order, derivative. To find the second derivative, dd𝑦𝑥, we first find the first derivative dd𝑦𝑥. Then, we can find the second derivative by differentiating the first derivative. Let’s consider an example where we find the second derivative from an implicit equation.

Example 4: Second Derivatives Using Implicit Differentiation

Given that 2𝑦5𝑥=4sincos, determine 𝑦 by implicit differentiation.

Answer

We begin by differentiating implicitly to find the first derivative as follows: 2(𝑦)𝑦𝑥+5𝑥=0.cosddsin

At this point, we can either differentiate again or first rearrange to make dd𝑦𝑥 the subject. Generally, when we differentiate again, we will get an expression for dd𝑦𝑥 in terms of dd𝑦𝑥. For this reason, it is often helpful to rearrange the equation first. By subtracting 5𝑥sin from both sides of the equation, we get 2(𝑦)𝑦𝑥=5𝑥.cosddsin

If cos𝑦0, we can divide by 2𝑦cos to get ddsincos𝑦𝑥=5𝑥2𝑦.

At this point, we can use the quotient rule, 𝑓𝑔=𝑓𝑔𝑓𝑔𝑔, to differentiate again to get ddcoscossinsincoscoscossinsincos𝑦𝑥=10𝑥𝑦+10𝑥𝑦4𝑦=5𝑥𝑦+5𝑥𝑦2𝑦.dddd

We can now substitute in the expression for dd𝑦𝑥 that we calculated earlier to get ddcoscossinsincos𝑦𝑥=5𝑥𝑦+5𝑥𝑦2𝑦.sincos

Expressing the numerator as a single fraction, we have ddcoscoscossinsincossinsincoscoscos𝑦𝑥=2𝑦=10𝑥𝑦25𝑥𝑦4𝑦=25𝑥𝑦10𝑥𝑦4𝑦.coscossinsinsincos

This formula is valid for all 𝑥 and 𝑦 such that cos𝑦0.

Let’s look at an example where we compute the third derivative, dd𝑦𝑥, from an implicit equation.

Example 5: Higher-Order Derivatives Using Implicit Differentiation

Find dd𝑦𝑥, given that 6𝑥+6𝑦=25.

Answer

We begin by differentiating 6𝑥+6𝑦=25 implicitly to get 12𝑥+12𝑦𝑦𝑥=0.dd

At this point, we can either differentiate again or first rearrange to make dd𝑦𝑥 the subject. Generally, when we differentiate again, we will get an expression for dd𝑦𝑥 in terms of dd𝑦𝑥. For this reason, it is often helpful to rearrange the equation first. Therefore, by subtracting 12𝑥 from both sides of the equation, we have 12𝑦𝑦𝑥=12𝑥.dd

Hence, when 𝑦0, dd𝑦𝑥=𝑥𝑦.

We can now use the quotient rule, 𝑓𝑔=𝑓𝑔𝑓𝑔𝑔, to differentiate again to get dd𝑦𝑥=𝑦𝑥𝑦.dd

Substituting in the expression for dd𝑦𝑥, we have dd𝑦𝑥=𝑦𝑥𝑦=𝑦+𝑥𝑦.

Since 6𝑥+6𝑦=25, we can rewrite this as dd𝑦𝑥=256𝑦=256𝑦.

We can now differentiate again to get an expression for dd𝑦𝑥 as follows: dddd𝑦𝑥=252𝑦𝑦𝑥.

Substituting dd𝑦𝑥=𝑥𝑦 into the equation yields dd𝑦𝑥=252𝑦𝑥𝑦=25𝑥2𝑦.

Let’s recap a few important concepts from this explainer.

Key Points

  • When given a function defined implicitly, we can differentiate each term of the equation using the following version of the chain rule for the terms including our dependent variable 𝑦: dddd𝑥(𝑓(𝑦))=𝑓(𝑦)𝑦𝑥.
  • Even when it is possible to rewrite a relation as an explicit function, it is often simpler to use implicit differentiation.
  • When we differentiate implicitly, in general, we will get an expression for dd𝑦𝑥 in terms of both 𝑥 and 𝑦.
  • We can find higher-order derivatives with implicit differentiation. In such cases, we will often need to substitute expressions for lower-order derivatives, and even the equation which defines the function implicitly, to simplify the expressions.

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