Explainer: Implicit Differentiation

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=√1βˆ’π‘₯,𝑦=𝑒π‘₯.οŠ¨ο—οŽ‘cos

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

However, this might not be possible in general. For example, consider the relation 𝑦𝑦=π‘₯π‘₯.sinsin

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

Chain Rule

Given a function β„Ž that is differentiable at π‘₯ and a function 𝑔 that is differentiable at β„Ž(π‘₯), their composition 𝑓=π‘”βˆ˜β„Ž defined by 𝑓(π‘₯)=𝑔(β„Ž(π‘₯)) is differentiable at π‘₯ and its derivative π‘“οŽ˜ is given by 𝑓(π‘₯)=β„Ž(π‘₯)𝑔(β„Ž(π‘₯)).

We can write this in Leibniz notation as dddddd𝑦π‘₯=𝑦𝑒⋅𝑒π‘₯, where 𝑦=𝑔(𝑒) and 𝑒=β„Ž(π‘₯).

We will demonstrate the method by example.

Example 1: Implicit Differentiation

Consider the equation π‘₯+𝑦=1.

  1. Using implicit differentiation, find an expression for dd𝑦π‘₯ in terms of π‘₯ and 𝑦.
  2. For the semicircle where 𝑦β‰₯0, express 𝑦 explicitly in terms of π‘₯; then, differentiate this expression to get an expression for dd𝑦π‘₯ in terms of π‘₯.

Answer

Part 1

To implicitly differentiate, we need to differentiate both sides of the equation with respect to π‘₯ as follows: ddddπ‘₯ο€Ήπ‘₯+𝑦=π‘₯(1).

Using the linearity of the derivative, we can rewrite this as ddddπ‘₯ο€Ήπ‘₯+π‘₯𝑦=0.

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

2π‘₯+π‘₯𝑦=0.dd(1)

As for the second, since π‘¦οŠ¨ is a function of 𝑦 and we can consider 𝑦 to be a function of π‘₯, we can use the chain rule to get ddddπ‘₯𝑦=2𝑦𝑦π‘₯.

Substituting this back into equation (1), we have 2π‘₯+2𝑦𝑦π‘₯=0.dd

We can now rearrange to make dd𝑦π‘₯ the subject. By subtracting 2π‘₯ from both sides, and dividing by two, we have 𝑦𝑦π‘₯=βˆ’π‘₯.dd

For all 𝑦≠0, we can rewrite this as dd𝑦π‘₯=βˆ’π‘₯𝑦.

Part 2

We begin by rearranging the equation we have been given to make 𝑦 the subject. By subtracting π‘₯ from both sides of the equation, we have 𝑦=1βˆ’π‘₯.

Since we are only interested in the semicircle where 𝑦β‰₯0, we can take the positive square root of both sides of the equation to get 𝑦=√1βˆ’π‘₯.

Now we have an explicit formula for 𝑦 in terms of π‘₯, 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 𝑒=1βˆ’π‘₯, we have 𝑦=βˆšπ‘’. We now differentiate to get dddd𝑦𝑒=12βˆšπ‘’,𝑒π‘₯=βˆ’2π‘₯.

Applying the chain rule, dddddd𝑦π‘₯=𝑦𝑒⋅𝑒π‘₯, we have dd𝑦π‘₯=12√1βˆ’π‘₯(βˆ’2π‘₯)=βˆ’π‘₯√1βˆ’π‘₯.

Notice that since 𝑦=√1βˆ’π‘₯, we could express this as dd𝑦π‘₯=βˆ’π‘₯𝑦, 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 dd𝑦π‘₯ in terms of both π‘₯ and 𝑦.
  • In general, when differentiating implicitly, we use the following version of the chain rule: ddddπ‘₯(𝑓(𝑦))=𝑓(𝑦)𝑦π‘₯.

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 π‘¦βˆ’24π‘₯+24π‘₯=0 describes a curve in the plane.

  1. Find the coordinates of two points on this curve, where π‘₯=βˆ’12.
  2. Determine the equation of the tangent at the points where π‘₯=βˆ’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 π‘₯=βˆ’12, we substitute this value into the equation and solve for 𝑦. Hence, 0=π‘¦βˆ’24ο€Όβˆ’12+24ο€Όβˆ’12=𝑦+3βˆ’12=π‘¦βˆ’9.

Hence, 𝑦=9, which implies 𝑦=3 and 𝑦=βˆ’3 are the two 𝑦-values where π‘₯=βˆ’12, which we can write as coordinates as ο€Όβˆ’12,3 and ο€Όβˆ’12,βˆ’3.

Part 2

We need to find the equation of the tangent to the curve at ο€Όβˆ’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 2𝑦𝑦π‘₯βˆ’72π‘₯+24=0.dd

Hence, 2𝑦𝑦π‘₯=72π‘₯βˆ’24.dd

Since, for the point we are interested in, 𝑦≠0, we can rewrite this as dd𝑦π‘₯=36π‘₯βˆ’12𝑦.

Substituting in π‘₯=βˆ’12 and 𝑦=3, we have dd𝑦π‘₯|||=36ο€»βˆ’ο‡βˆ’123=124βˆ’4=βˆ’1.(ο—οŽ•ο˜)(οŠ±οŽ•οŠ©)

Therefore, the slope of the tangent at ο€Όβˆ’12,3 is βˆ’1. Hence, using the slope–point form of the equation of a line, π‘¦βˆ’π‘¦=π‘š(π‘₯βˆ’π‘₯), we can find the equation of the tangent as follows: π‘¦βˆ’3=βˆ’1ο€Όπ‘₯βˆ’ο€Όβˆ’12.

We can simplify this to 𝑦=52βˆ’π‘₯.

Part 3

To find the coordinates of another point where this tangent crosses the curve, we can substitute 𝑦=52βˆ’π‘₯ into the equation for the curve and try to solve for π‘₯. We can then find the corresponding value of 𝑦. Therefore, ο€Ό52βˆ’π‘₯οˆβˆ’24π‘₯+24π‘₯=0.

Expanding the parentheses, we have 254βˆ’5π‘₯+π‘₯βˆ’24π‘₯+24π‘₯=0.

Hence, 24π‘₯βˆ’π‘₯βˆ’19π‘₯βˆ’254=0.

We already know that the tangent passes through the point ο€Όβˆ’12,3. Hence, π‘₯=βˆ’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 ο€Όπ‘₯+12οˆο€Ήπ‘Žπ‘₯+𝑏π‘₯+𝑐=24π‘₯βˆ’π‘₯βˆ’19π‘₯βˆ’254.

We can see that the first term in the parentheses must be 24π‘₯ to ensure we get the correct coefficient for the π‘₯-term. Hence, ο€Όπ‘₯+12οˆο€Ή24π‘₯+𝑏π‘₯+𝑐=24π‘₯βˆ’π‘₯βˆ’19π‘₯βˆ’254.

Similarly, we can see that the constant term in the parentheses must be 252. Therefore, ο€Όπ‘₯+12οˆο€Ό24π‘₯+𝑏π‘₯βˆ’252=24π‘₯βˆ’π‘₯βˆ’19π‘₯βˆ’254.

Finally, we need to find the middle term in the parentheses. We can do this by comparing the coefficients of the π‘₯-term on both sides of the equation. On the right-hand side, we have a coefficient of βˆ’1, whereas on the left-hand side, we have a coefficient of 𝑏+12. Therefore, 𝑏=βˆ’13 and ο€Όπ‘₯+12οˆο€Ό24π‘₯βˆ’13π‘₯βˆ’252=24π‘₯βˆ’π‘₯βˆ’19π‘₯βˆ’254.

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 βˆ’19π‘₯. Therefore, considering the coefficient of π‘₯ on the left-hand side, we have βˆ’132βˆ’252=βˆ’19 as required.

We now have a quadratic equation we can solve. Using the quadratic formula or otherwise, we find this has solutions π‘₯=βˆ’12 and π‘₯=2524. Therefore, the tangent intersects the curve again at the point where π‘₯=2524. Using the equation of the tangent, we can substitute in this value of π‘₯ to find the corresponding value of 𝑦 as follows: 𝑦=52βˆ’2524=3524.

Hence, the tangent at ο€Όβˆ’12,3 meets the curve again at the point ο€Ό2524,3524.

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<π‘₯<πœ‹, the tangent to 9𝑦=(5π‘₯+3𝑦)cos that has slope βˆ’512, giving your equation in terms of πœ‹.

Answer

We have been given the slope of a tangent and would like to find the values of π‘₯ 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 dd𝑦π‘₯. Using implicit differentiation, 9𝑦π‘₯=βˆ’ο€½5+3𝑦π‘₯(5π‘₯+3𝑦).ddddsin

Since the derivative is equal to the slope of the tangent, we can substitute βˆ’512 in for dd𝑦π‘₯ as follows: 9ο€Όβˆ’512=βˆ’ο€Ό5+3ο€Όβˆ’512(5π‘₯+3𝑦).sin

Simplifying, we get βˆ’154=βˆ’154(5π‘₯+3𝑦)1=(5π‘₯+3𝑦).sinsin

Hence, 5π‘₯+3𝑦=πœ‹2.

Sometimes, we are interested in higher-order derivatives, in these cases we can also use implicit differentiation.

Example 4: Second Derivatives Using Implicit Differentiation

Given that 2π‘¦βˆ’5π‘₯=βˆ’4sincos, determine π‘¦οŽ™ by implicit differentiation.

Answer

We begin by differentiating implicitly to find the first derivative as follows: 2(𝑦)𝑦π‘₯+5π‘₯=0.cosddsin

At this point, we can either differentiate again or first rearrange to make dd𝑦π‘₯ the subject. Generally, when we differentiate again, we will get an expression for ddοŠ¨οŠ¨π‘¦π‘₯ in terms of dd𝑦π‘₯. For this reason, it is often helpful to rearrange the equation first. By subtracting 5π‘₯sin from both sides of the equation, we get 2(𝑦)𝑦π‘₯=βˆ’5π‘₯.cosddsin

If cos𝑦≠0, we can divide by 2𝑦cos to get ddsincos𝑦π‘₯=βˆ’5π‘₯2𝑦.

At this point, we can use the quotient rule, 𝑓𝑔=π‘“π‘”βˆ’π‘“π‘”π‘”, to differentiate again to get ddcoscossinsincoscoscossinsincosοŠ¨οŠ¨ο˜ο—οŠ¨ο˜ο—οŠ¨π‘¦π‘₯=βˆ’10π‘₯𝑦+10π‘₯𝑦4𝑦=βˆ’5π‘₯𝑦+5π‘₯𝑦2𝑦.dddd

We can now substitute in the expression for dd𝑦π‘₯ that we calculated earlier to get ddcoscossinsincosοŠ¨οŠ¨οŠ«ο—οŠ¨ο˜οŠ¨π‘¦π‘₯=βˆ’5π‘₯𝑦+5π‘₯π‘¦ο€Όβˆ’οˆ2𝑦.sincos

Expressing the numerator as a single fraction, we have ddcoscoscossinsincossinsincoscoscosοŠ¨οŠ¨οŠ§οŠ¦ο—ο˜οŠ±οŠ¨οŠ«ο—ο˜ο—οŠ¨ο˜οŠ¨οŠ¨οŠ¨οŠ©οŠ¨οŠ¨οŠ©π‘¦π‘₯=βˆ’2𝑦=βˆ’10π‘₯π‘¦βˆ’25π‘₯𝑦4𝑦=25π‘₯π‘¦βˆ’10π‘₯𝑦4𝑦.coscossinsinsincos

This formula is valid for all π‘₯ and 𝑦 such that cos𝑦≠0.

Example 5: Higher-Order Derivatives Using Implicit Differentiation

Find ddοŠ©οŠ©π‘¦π‘₯, given that 6π‘₯+6𝑦=25.

Answer

We being by differentiating 6π‘₯+6𝑦=25 implicitly to get 12π‘₯+12𝑦𝑦π‘₯=0.dd

At this point, we can either differentiate again or first rearrange to make dd𝑦π‘₯ the subject. Generally, when we differentiate again, we will get an expression for ddοŠ¨οŠ¨π‘¦π‘₯ in terms of dd𝑦π‘₯. For this reason, it is often helpful to rearrange the equation first. Therefore, by subtracting 12π‘₯ from both sides of the equation, we have 12𝑦𝑦π‘₯=βˆ’12π‘₯.dd

Hence, when 𝑦≠0, dd𝑦π‘₯=βˆ’π‘₯𝑦.

We can now use the quotient rule, 𝑓𝑔=π‘“π‘”βˆ’π‘“π‘”π‘”, to differentiate again to get ddοŠ¨οŠ¨ο˜ο—οŠ¨π‘¦π‘₯=βˆ’π‘¦βˆ’π‘₯𝑦.dd

Substituting in the expression for dd𝑦π‘₯, we have ddοŠ¨οŠ¨ο—ο˜οŠ¨οŠ¨οŠ¨οŠ©π‘¦π‘₯=βˆ’π‘¦βˆ’π‘₯ο€Όβˆ’οˆπ‘¦=βˆ’π‘¦+π‘₯𝑦.

Since 6π‘₯+6𝑦=25, we can rewrite this as ddοŠ¨οŠ¨οŠ©οŠ±οŠ©π‘¦π‘₯=βˆ’256𝑦=βˆ’256𝑦.

We can now differentiate again to get an expression for ddοŠ©οŠ©π‘¦π‘₯ as follows: ddddοŠͺ𝑦π‘₯=252𝑦𝑦π‘₯.

Substituting dd𝑦π‘₯=βˆ’π‘₯𝑦 into the equation yields ddοŠͺοŠ«π‘¦π‘₯=252π‘¦ο€½βˆ’π‘₯𝑦=βˆ’25π‘₯2𝑦.

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: ddddπ‘₯(𝑓(𝑦))=𝑓(𝑦)𝑦π‘₯.
  • 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 dd𝑦π‘₯ in terms of both π‘₯ 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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