4.1 Implicit Differentiationap Calculus

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A mathematical constant is a key number whose value is fixed by an unambiguous definition, often referred to by a symbol (e.g., an alphabet letter), or by mathematicians' names to facilitate using it across multiple mathematical problems.

In the previous sections we learned to find the derivative, ( frac{dy}{dx}), or (y^prime ), when (y) is given explicitly as a function of (x). That is, if we know (y=f(x)) for some function (f), we can find (y^prime ). For example, given (y=3x^2-7), we can easily find (y^prime =6x). (Here we explicitly state how (x) and (y) are related. Knowing (x), we can directly find (y).)

Sometimes the relationship between (y) and (x) is not explicit; rather, it is implicit. For instance, we might know that (x^2-y=4). This equality defines a relationship between (x) and (y); if we know (x), we could figure out (y). Can we still find (y^prime )? In this case, sure; we solve for (y) to get (y=x^2-4) (hence we now know (y) explicitly) and then differentiate to get (y^prime =2x).

Notes 4.1 Video When to use implicit differentiation (Day 1) Notes 4.1 Video How to use implicit differentiation (Day 1) Notes 4.1 Video AP-style type questions (Day 1). Implicit Differentiation. Finding the derivative when you can’t solve for y. You may like to read Introduction to Derivatives and Derivative Rules first. Implicit vs Explicit. A function can be explicit or implicit: Explicit: 'y = some function of x'. When we know x we can calculate y directly. Implicit: 'some function of y and x equals. This lesson contains the following Essential Knowledge (EK) concepts for the.AP Calculus course.Click here for an overview of all the EK's in this course. EK 2.1C5. AP® is a trademark registered and owned by the College Board, which was not involved in the production of, and does not endorse, this site.

Sometimes the implicit relationship between (x) and (y) is complicated. Suppose we are given (sin(y)+y^3=6-x^3). A graph of this implicit function is given in Figure 2.19. In this case there is absolutely no way to solve for (y) in terms of elementary functions. The surprising thing is, however, that we can still find (y^prime ) via a process known as implicit differentiation.

Implicit differentiation is a technique based on the Chain Rule that is used to find a derivative when the relationship between the variables is given implicitly rather than explicitly (solved for one variable in terms of the other).

We begin by reviewing the Chain Rule. Let (f) and (g) be functions of (x). Then [frac{d}{dx}Big(f(g(x))Big) = f^prime(g(x))cdot g'(x).] Suppose now that (y=g(x)). We can rewrite the above as [frac{d}{dx}Big(f(y))Big) = f^prime(y))cdot y^prime , quad text{or}quad frac{d}{dx}Big(f(y))Big)= f^prime(y)cdot frac{dy}{dx}.label{2.1}tag{2.1}] These equations look strange; the key concept to learn here is that we can find (y^prime ) even if we don't exactly know how (y) and (x) relate.

We demonstrate this process in the following example.

Example 67: Using Implicit Differentiation

Find (y^prime ) given that (sin(y) + y^3=6-x^3).

Solution

We start by taking the derivative of both sides (thus maintaining the equality.) We have :

[ frac{d}{dx}Big(sin(y) + y^3Big)=frac{d}{dx}Big(6-x^3Big).]

The right hand side is easy; it returns (-3x^2).

The left hand side requires more consideration. We take the derivative term--by--term. Using the technique derived from Equation 2.1 above, we can see that [frac{d}{dx}Big(sin yBig) = cos y cdot y^prime .]

We apply the same process to the (y^3) term.

[frac{d}{dx}Big(y^3Big) = frac{d}{dx}Big((y)^3Big) = 3(y)^2cdot y^prime .]

Putting this together with the right hand side, we have

[cos(y)y^prime +3y^2y^prime = -3x^2.]

Now solve for (y^prime ).

[begin{align*} cos(y)y^prime +3y^2y^prime &= -3x^2. big(cos y+3y^2big)y^prime &= -3x^2 y^prime &= frac{-3x^2}{cos y+3y^2} end{align*}]

This equation for (y^prime ) probably seems unusual for it contains both (x) and (y) terms. How is it to be used? We'll address that next.

Implicit functions are generally harder to deal with than explicit functions. With an explicit function, given an (x) value, we have an explicit formula for computing the corresponding (y) value. With an implicit function, one often has to find (x) and (y) values at the same time that satisfy the equation. It is much easier to demonstrate that a given point satisfies the equation than to actually find such a point.

For instance, we can affirm easily that the point ((sqrt[3]{6},0)) lies on the graph of the implicit function (sin y + y^3=6-x^3). Plugging in (0) for (y), we see the left hand side is (0). Setting (x=sqrt[3]6), we see the right hand side is also (0); the equation is satisfied. The following example finds the equation of the tangent line to this function at this point.

Example 68: Using Implicit Differentiation to find a tangent line

Find the equation of the line tangent to the curve of the implicitly defined function (sin y + y^3=6-x^3) at the point ((sqrt[3]6,0)).

Solution

In Example 67 we found that [y^prime = frac{-3x^2}{cos y +3y^2}.]We find the slope of the tangent line at the point ((sqrt[3]6,0)) by substituting (sqrt[3]6) for (x) and (0) for (y). Thus at the point ((sqrt[3]6,0)), we have the slope as [y^prime = frac{-3(sqrt[3]{6})^2}{cos 0 + 3cdot0^2} = frac{-3sqrt[3]{36}}{1} approx -9.91.]

Therefore the equation of the tangent line to the implicitly defined function (sin y + y^3=6-x^3) at the point ((sqrt[3]{6},0)) is [y = -3sqrt[3]{36}(x-sqrt[3]{6})+0 approx -9.91x+18.]The curve and this tangent line are shown in Figure 2.20.

This suggests a general method for implicit differentiation. For the steps below assume (y) is a function of (x).

1. Take the derivative of each term in the equation. Treat the (x) terms like normal. When taking the derivatives of (y) terms, the usual rules apply except that, because of the Chain Rule, we need to multiply each term by (y^prime ).
2. Get all the (y^prime ) terms on one side of the equal sign and put the remaining terms on the other side.
3. Factor out (y^prime ); solve for (y^prime ) by dividing.

Practical Note: When working by hand, it may be beneficial to use the symbol (frac{dy}{dx}) instead of (y^prime ), as the latter can be easily confused for (y) or (y^1).

Example 69: Using Implicit Differentiation

Given the implicitly defined function (y^3+x^2y^4=1+2x), find (y^prime ).

Solution

We will take the implicit derivatives term by term. The derivative of (y^3) is (3y^2y^prime ).

The second term, (x^2y^4), is a little tricky. It requires the Product Rule as it is the product of two functions of (x): (x^2) and (y^4). Its derivative is (x^2(4y^3y^prime ) + 2xy^4). The first part of this expression requires a (y^prime ) because we are taking the derivative of a (y) term. The second part does not require it because we are taking the derivative of (x^2).

The derivative of the right hand side is easily found to be (2). In all, we get:

[3y^2y^prime + 4x^2y^3y^prime + 2xy^4 = 2.]

Move terms around so that the left side consists only of the (y^prime ) terms and the right side consists of all the other terms:

[3y^2y^prime + 4x^2y^3y^prime = 2-2xy^4.]

Factor out (y^prime ) from the left side and solve to get

[y^prime = frac{2-2xy^4}{3y^2+4x^2y^3}.]

To confirm the validity of our work, let's find the equation of a tangent line to this function at a point. It is easy to confirm that the point ((0,1)) lies on the graph of this function. At this point, (y^prime = 2/3). So the equation of the tangent line is (y = 2/3(x-0)+1). The function and its tangent line are graphed in Figure 2.21.

Notice how our function looks much different than other functions we have seen. For one, it fails the vertical line test. Such functions are important in many areas of mathematics, so developing tools to deal with them is also important.

Example 70: Using Implicit Differentiation

Given the implicitly defined function (sin(x^2y^2)+y^3=x+y), find (y^prime ).

Solution

Differentiating term by term, we find the most difficulty in the first term. It requires both the Chain and Product Rules.

[begin{align*} frac{d}{dx}Big(sin(x^2y^2)Big) &= cos(x^2y^2)cdotfrac{d}{dx}Big(x^2y^2Big) &= cos(x^2y^2)cdotbig(x^2(2yy^prime )+2xy^2big) &= 2(x^2yy^prime +xy^2)cos(x^2y^2). end{align*} ]

We leave the derivatives of the other terms to the reader. After taking the derivatives of both sides, we have

[2(x^2yy^prime +xy^2)cos(x^2y^2) + 3y^2y^prime = 1 + y^prime .]

We now have to be careful to properly solve for (y^prime ), particularly because of the product on the left. It is best to multiply out the product. Doing this, we get

[2x^2ycos(x^2y^2)y^prime + 2xy^2cos(x^2y^2) + 3y^2y^prime = 1 + y^prime .]

From here we can safely move around terms to get the following:

[2x^2ycos(x^2y^2)y^prime + 3y^2y^prime - y^prime = 1 - 2xy^2cos(x^2y^2).]

Then we can solve for (y^prime ) to get

[y^prime = frac{1 - 2xy^2cos(x^2y^2)}{2x^2ycos(x^2y^2)+3y^2-1}.]

A graph of this implicit function is given in Figure 2.22. It is easy to verify that the points ((0,0)), ((0,1)) and ((0,-1)) all lie on the graph. We can find the slopes of the tangent lines at each of these points using our formula for (y^prime ).

At ((0,0)), the slope is (-1).

At ((0,1)), the slope is (1/2).

At ((0,-1)), the slope is also (1/2).

The tangent lines have been added to the graph of the function in Figure 2.23.

Quite a few 'famous' curves have equations that are given implicitly. We can use implicit differentiation to find the slope at various points on those curves. We investigate two such curves in the next examples.

Example 71: Finding slopes of tangent lines to a circle

Find the slope of the tangent line to the circle (x^2+y^2=1) at the point ((1/2, sqrt{3}/2)).

Solution

Taking derivatives, we get (2x+2yy^prime =0). Solving for (y^prime ) gives: [ y^prime = frac{-x}{y}.]

This is a clever formula. Recall that the slope of the line through the origin and the point ((x,y)) on the circle will be (y/x). We have found that the slope of the tangent line to the circle at that point is the opposite reciprocal of (y/x), namely, (-x/y). Hence these two lines are always perpendicular.

At the point ((1/2, sqrt{3}/2)), we have the tangent line's slope as

[y^prime = frac{-1/2}{sqrt{3}/2} = frac{-1}{sqrt{3}} approx -0.577.]

A graph of the circle and its tangent line at ((1/2,sqrt{3}/2)) is given in Figure 2.24, along with a thin dashed line from the origin that is perpendicular to the tangent line. (It turns out that all normal lines to a circle pass through the center of the circle.)

This section has shown how to find the derivatives of implicitly defined functions, whose graphs include a wide variety of interesting and unusual shapes. Implicit differentiation can also be used to further our understanding of 'regular' differentiation.

One hole in our current understanding of derivatives is this: what is the derivative of the square root function? That is, [frac{d}{dx}big(sqrt{x}big) = frac{d}{dx}big(x^{1/2}big) = text{?}]

We allude to a possible solution, as we can write the square root function as a power function with a rational (or, fractional) power. We are then tempted to apply the Power Rule and obtain [frac{d}{dx}big(x^{1/2}big) = frac12x^{-1/2} = frac{1}{2sqrt{x}}.]

The trouble with this is that the Power Rule was initially defined only for positive integer powers, (n>0). While we did not justify this at the time, generally the Power Rule is proved using something called the Binomial Theorem, which deals only with positive integers. The Quotient Rule allowed us to extend the Power Rule to negative integer powers. Implicit Differentiation allows us to extend the Power Rule to rational powers, as shown below.

Let (y = x^{m/n}), where (m) and (n) are integers with no common factors (so (m=2) and (n=5) is fine, but (m=2) and (n=4) is not). We can rewrite this explicit function implicitly as (y^n = x^m). Now apply implicit differentiation.

[begin{align*}y &= x^{m/n} y^n &= x^m frac{d}{dx}big(y^nbig) &= frac{d}{dx}big(x^mbig) ncdot y^{n-1}cdot y^prime &= mcdot x^{m-1} y^prime &= frac{m}{n} frac{x^{m-1}}{y^{n-1}} quad text{(now substitute (x^{m/n}) for (y))} &= frac{m}{n} frac{x^{m-1}}{(x^{m/n})^{n-1}} quad text{(apply lots of algebra)} &= frac{m}n x^{(m-n)/n} &= frac{m}n x^{m/n -1}.end{align*}]

The above derivation is the key to the proof extending the Power Rule to rational powers. Using limits, we can extend this once more to include all powers, including irrational (even transcendental!) powers, giving the following theorem.

Theorem 21: Power Rule for Differentiation

Let (f(x) = x^n), where (nneq 0) is a real number. Then (f) is a differentiable function, and (f^prime(x) = ncdot x^{n-1}).

This theorem allows us to say the derivative of (x^pi) is (pi x^{pi -1}).

We now apply this final version of the Power Rule in the next example, the second investigation of a 'famous' curve.

Example 72: Using the Power Rule

Find the slope of (x^{2/3}+y^{2/3}=8) at the point ((8,8)).

Solution

This is a particularly interesting curve called an astroid. It is the shape traced out by a point on the edge of a circle that is rolling around inside of a larger circle, as shown in Figure 2.25.

To find the slope of the astroid at the point ((8,8)), we take the derivative implicitly.

[begin{align*} frac23x^{-1/3}+frac23y^{-1/3}y^prime &=0 frac23y^{-1/3}y^prime &= -frac23x^{-1/3} y^prime &= -frac{x^{-1/3}}{y^{-1/3}} y^prime &= -frac{y^{1/3}}{x^{1/3}} = -sqrt[3]{frac{y}x}. end{align*}]

Plugging in (x=8) and (y=8), we get a slope of (-1). The astroid, with its tangent line at ((8,8)), is shown in Figure 2.26.

Implicit Differentiation and the Second Derivative

We can use implicit differentiation to find higher order derivatives. In theory, this is simple: first find (frac{dy}{dx}), then take its derivative with respect to (x). In practice, it is not hard, but it often requires a bit of algebra. We demonstrate this in an example.

Example 73: Finding the second derivative

Given (x^2+y^2=1), find (frac{d^2y}{dx^2} = y^{primeprime}).

Solution

We found that (y^prime = frac{dy}{dx} = -x/y) in Example 71. To find (y^{primeprime}), we apply implicit differentiation to (y^prime ).

[begin{align*} y^{primeprime} &= frac{d}{dx}big(y^prime big) &= frac{d}{dx}left(-frac xyright)qquad text{(Now use the Quotient Rule.)} &= -frac{y(1) - x(y^prime )}{y^2} end{align*}]

replace (y^prime ) with (-x/y):

[begin{align*}&= -frac{y-x(-x/y)}{y^2} &= -frac{y+x^2/y}{y^2}.end{align*}]

While this is not a particularly simple expression, it is usable. We can see that (y^{primeprime}>0) when (y<0) and (y^{primeprime}<0) when (y>0). In Section 3.4, we will see how this relates to the shape of the graph.

Logarithmic Differentiation

Consider the function (y=x^x); it is graphed in Figure 2.27. It is well--defined for (x>0) and we might be interested in finding equations of lines tangent and normal to its graph. How do we take its derivative?

The function is not a power function: it has a 'power' of (x), not a constant. It is not an exponential function: it has a 'base' of (x), not a constant.

A differentiation technique known as logarithmic differentiation becomes useful here. The basic principle is this: take the natural log of both sides of an equation (y=f(x)), then use implicit differentiation to find (y^prime ). We demonstrate this in the following example.

Example 74: Using Logarithmic Differentiation

Given (y=x^x), use logarithmic differentiation to find (y^prime ).

Solution

As suggested above, we start by taking the natural log of both sides then applying implicit differentiation.

[begin{align*} y &= x^x ln (y) &= ln (x^x) text{(apply logarithm rule)} ln (y) &= xln x text{(now use implicit differentiation)} frac{d}{dx}Big(ln (y)Big) &= frac{d}{dx}Big(xln xBig) frac{y^prime }{y} &= ln x + xcdotfrac1x frac{y^prime }{y} &= ln x + 1 y^prime &= ybig(ln x+1big) text{(substitute (y=x^x))} y^prime &= x^xbig(ln x+1big). end{align*} ]

To 'test' our answer, let's use it to find the equation of the tangent line at (x=1.5). The point on the graph our tangent line must pass through is ((1.5, 1.5^{1.5}) approx (1.5, 1.837)). Using the equation for (y^prime ), we find the slope as

[y^prime = 1.5^{1.5}big(ln 1.5+1big) approx 1.837(1.405) approx 2.582.]

Thus the equation of the tangent line is (y = 1.6833(x-1.5)+1.837). Figure 2.28 graphs (y=x^x) along with this tangent line.

Implicit differentiation proves to be useful as it allows us to find the instantaneous rates of change of a variety of functions. In particular, it extended the Power Rule to rational exponents, which we then extended to all real numbers. In the next section, implicit differentiation will be used to find the derivatives of inverse functions, such as (y=sin^{-1} x).

Finding the derivative when you can’t solve for y

You may like to read Introduction to Derivatives and Derivative Rules first.

Implicit vs Explicit

A function can be explicit or implicit:

Explicit: 'y = some function of x'. When we know x we can calculate y directly.

Implicit: 'some function of y and x equals something else'. Knowing x does not lead directly to y.

Example: A Circle

The graph of x² + y² = 3²

How to do Implicit Differentiation

• Differentiate with respect to x
• Collect all the dydx on one side
• Solve for dydx

Example: x² + y² = r²

Differentiate with respect to x:

ddx(x²) + ddx(y²) = ddx(r²)

Let's solve each term:

Which gives us:

2x + 2ydydx = 0

Collect all the dydx on one side

ydydx = −x

Solve for dydx:

dydx = −xy

The Chain Rule Using dydx

Let's look more closely at how ddx(y²) becomes 2ydydx

The Chain Rule says:

dudx = dudydydx

Substitute in u = y²:

ddx(y²) = ddy(y²)dydx

And then:

ddx(y²) = 2ydydx

Basically, all we did was differentiate with respect to y and multiply by dydx

Another common notation is to use ’ to mean ddx

The Chain Rule Using ’

The Chain Rule can also be written using ’ notation:

f(g(x))’ = f’(g(x))g’(x)

g(x) is our function 'y', so:

f(y)’ = f’(y)y’

f(y) = y², so f’(y) = 2y:

f(y)’ = 2yy’

or alternatively: f(y)’ = 2y dydx

Again, all we did was differentiate with respect to y and multiply by dydx

Explicit

Let's also find the derivative using the explicit form of the equation.

• To solve this explicitly, we can solve the equation for y
• Then differentiate
• Then substitute the equation for y again

Example: x² + y² = r²

We get the same result this way!

You can try taking the derivative of the negative term yourself.

Chain Rule Again!

Yes, we used the Chain Rule again. Like this (note different letters, but same rule):

dydx = dydfdfdx

Substitute in f = (r² − x²):

ddx(f^½) = ddf(f^½)ddx(r² − x²)

4.1 Implicit Differentiationap Calculus Multiple Choice

Derivatives:

ddx(f^½) = ½(f^−½) (−2x)

And substitute back f = (r² − x²):

ddx(r² − x²)^½ = ½((r² − x²)^−½) (−2x)

And we simplified from there.

4.1 Implicit Differentiationap Calculus Calculator

Using The Derivative

OK, so why find the derivative y’ = −x/y ?

Well, for example, we can find the slope of a tangent line.

Example: what is the slope of a circle centered at the origin with a radius of 5 at the point (3,4)?

No problem, just substitute it into our equation:

dydx = −x/y

dydx = −3/4

And for bonus, the equation for the tangent line is:

y = −3/4 x + 25/4

Another Example

Sometimes the implicit way works where the explicit way is hard or impossible.

Example: 10x⁴ - 18xy² + 10y³ = 48

How do we solve for y? We don't have to!

• First, differentiate with respect to x (use the Product Rule for the xy² term).
• Then move all dy/dx terms to the left side.
• Solve for dy/dx

Like this:

(the middle term is explained below)

And we get:

Product Rule

For the middle term we used the Product Rule: (fg)’ = f g’ + f’ g

Because (y²)’ = 2ydydx(we worked that out in a previous example)

Oh, and dxdx = 1, in other words x’ = 1

Inverse Functions

Implicit differentiation can help us solve inverse functions.

The general pattern is:

• Start with the inverse equation in explicit form. Example: y = sin⁻¹(x)
• Rewrite it in non-inverse mode: Example: x = sin(y)
• Differentiate this function with respect to x on both sides.
• Solve for dy/dx

As a final step we can try to simplify more by substituting the original equation.

An example will help:

Example: the inverse sine function y = sin⁻¹(x)

We can also go one step further using the Pythagorean identity:

sin² y + cos² y = 1

cos y = √(1 − sin² y )

And, because sin(y) = x (from above!), we get:

cos y = √(1 − x²)

Which leads to:

dydx= 1√(1 − x²)

Example: the derivative of square root √x

Note: this is the same answer we get using the Power Rule:

Summary

• To Implicitly derive a function (useful when a function can't easily be solved for y)
  • Differentiate with respect to x
  • Collect all the dy/dx on one side
  • Solve for dy/dx
• To derive an inverse function, restate it without the inverse then use Implicit differentiation