2 Derivatives 2.5 The Chain Rule 3 The Graphical Behavior of Functions

2.6 Implicit Differentiation

In the previous sections we learned to find the derivative, $\frac{\operatorname{d}\!y}{\operatorname{d}\!x}$ , or $y\mkern 1.35mu ^{\prime}$ , 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 $y\mkern 1.35mu ^{\prime}$ . For example, given $y=3x^{2}-7$ , we can easily find $y\mkern 1.35mu ^{\prime}=6x$ . (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 $y\mkern 1.35mu ^{\prime}$ ? 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 $y\mkern 1.35mu ^{\prime}=2x$ .

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 equation is given in Figure 2.6.1. 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 $y\mkern 1.35mu ^{\prime}$ via a process known as implicit differentiation.

^†^†margin: Figure 2.6.1: A graph of the equation

\sin(y)+y^{3}=6-x^{3}

. Λ

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{\operatorname{d}\!}{\operatorname{d}\!x}\Bigl{(}f(g(x))\Bigr{)}=f\,^{% \prime}(g(x))\cdot g^{\prime}(x).

Suppose now that $y=g(x)$ . We can rewrite the above as

\frac{\operatorname{d}\!}{\operatorname{d}\!x}\Bigl{(}f(y)\Bigr{)}=f\,^{\prime% }(y)\cdot y\mkern 1.35mu ^{\prime},\quad\text{or}\quad\frac{\operatorname{d}\!% }{\operatorname{d}\!x}\Bigl{(}f(y)\Bigr{)}=f\,^{\prime}(y)\cdot\frac{% \operatorname{d}\!y}{\operatorname{d}\!x}.

(2.6.1)

These equations look strange; the key concept to learn here is that we can find $y\mkern 1.35mu ^{\prime}$ even if we don’t exactly know how $y$ and $x$ relate.

Watch the video:
Showing explicit and implicit differentiation give same result from https://youtu.be/2CsQ_l1S2_Y

We demonstrate this process in the following example.

Example 2.6.1 Using Implicit Differentiation

Find $y\mkern 1.35mu ^{\prime}$ given that $\sin(y)+y^{3}=6-x^{3}$ .

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

\frac{\operatorname{d}\!}{\operatorname{d}\!x}\Bigl{(}\sin(y)+y^{3}\Bigr{)}=% \frac{\operatorname{d}\!}{\operatorname{d}\!x}\Bigl{(}6-x^{3}\Bigr{)}.

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 (2.6.1) above, we can see that

\frac{\operatorname{d}\!}{\operatorname{d}\!x}\Bigl{(}\sin y\Bigr{)}=\cos y% \cdot y\mkern 1.35mu ^{\prime}.

We apply the same process to the $y^{3}$ term.

\frac{\operatorname{d}\!}{\operatorname{d}\!x}\Bigl{(}y^{3}\Bigr{)}=\frac{% \operatorname{d}\!}{\operatorname{d}\!x}\Bigl{(}(y)^{3}\Bigr{)}=3(y)^{2}\cdot y% \mkern 1.35mu ^{\prime}.

Putting this together with the right hand side, we have

\cos(y)y\mkern 1.35mu ^{\prime}+3y^{2}y\mkern 1.35mu ^{\prime}=-3x^{2}.

Now solve for $y\mkern 1.35mu ^{\prime}$ .

	$\displaystyle\cos(y)y\mkern 1.35mu ^{\prime}+3y^{2}y\mkern 1.35mu ^{\prime}$	$\displaystyle=-3x^{2}.$
	$\displaystyle\bigl{(}\cos y+3y^{2}\bigr{)}y\mkern 1.35mu ^{\prime}$	$\displaystyle=-3x^{2}$
	$\displaystyle y\mkern 1.35mu ^{\prime}$	$\displaystyle=\frac{-3x^{2}}{\cos y+3y^{2}}$

This equation for $y\mkern 1.35mu ^{\prime}$ 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 equation $\sin y+y^{3}=6-x^{3}$ . 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 an equation of the tangent line to this equation at this point.

Example 2.6.2 Using Implicit Differentiation to find a tangent line

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

SolutionIn Example 2.6.1 we found that

y\mkern 1.35mu ^{\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\mkern 1.35mu ^{\prime}=\frac{-3(\sqrt[3]{6})^{2}}{\cos 0+3\cdot 0^{2}}=\frac% {-3\sqrt[3]{36}}{1}\approx-9.91.

^†^†margin: Figure 2.6.2: The equation

\sin y+y^{3}=6-x^{3}

and its tangent line at the point

(\sqrt[3]{6},0)

. Λ

Therefore an equation of the tangent line to the implicitly defined curve $\sin y+y^{3}=6-x^{3}$ at the point $(\sqrt[3]{6},0)$ is

y=-3\sqrt[3]{36}(x-\sqrt[3]{6})+0\approx-9.91x+18.

The curve and this tangent line are shown in Figure 2.6.2.

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

(a)

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 $y\mkern 1.35mu ^{\prime}$ .
(b)

Get all the $y\mkern 1.35mu ^{\prime}$ terms on one side of the equal sign and put the remaining terms on the other side.
(c)

Factor out $y\mkern 1.35mu ^{\prime}$ ; solve for $y\mkern 1.35mu ^{\prime}$ by dividing.

Practical Note: When working by hand, it may be beneficial to use the symbol $\frac{\operatorname{d}\!y}{\operatorname{d}\!x}$ instead of $y\mkern 1.35mu ^{\prime}$ , as the latter can be easily confused for $y$ or $y^{1}$ .

Example 2.6.3 Using Implicit Differentiation

Given the implicitly defined function $y^{3}+x^{2}y^{4}=1+2x$ , find $y\mkern 1.35mu ^{\prime}$ .

SolutionWe will take the implicit derivatives term by term. Using the Chain Rule the derivative of $y^{3}$ is $3y^{2}y\mkern 1.35mu ^{\prime}$ .

The second term, $x^{2}y^{4}$ is a little more work. It requires the Product Rule as it is the product of two functions of $x$ : $x^{2}$ and $y^{4}$ . We see that $\dfrac{\operatorname{d}\!}{\operatorname{d}\!x}(x^{2}y^{4})$ is

	$\displaystyle x^{2}\cdot\frac{\operatorname{d}\!}{\operatorname{d}\!x}(y^{4})+% \frac{\operatorname{d}\!}{\operatorname{d}\!x}(x^{2})\cdot y^{4}$
	$\displaystyle x^{2}\cdot(4y^{3}y^{\prime})+2x\cdot y^{4}$

The first part of this expression requires a $y\mkern 1.35mu ^{\prime}$ 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 of the equation is found to be $2$ . In all, we get:

3y^{2}y\mkern 1.35mu ^{\prime}+4x^{2}y^{3}y\mkern 1.35mu ^{\prime}+2xy^{4}=2.

Move terms around so that the left side consists only of the $y\mkern 1.35mu ^{\prime}$ terms and the right side consists of all the other terms:

3y^{2}y\mkern 1.35mu ^{\prime}+4x^{2}y^{3}y\mkern 1.35mu ^{\prime}=2-2xy^{4}.

Factor out $y\mkern 1.35mu ^{\prime}$ from the left side and solve to get

y\mkern 1.35mu ^{\prime}=\frac{2-2xy^{4}}{3y^{2}+4x^{2}y^{3}}.

To confirm the validity of our work, let’s find the equation of a tangent line to this curve at a point. It is easy to confirm that the point $(0,1)$ lies on the graph of this curve. At this point, $y\mkern 1.35mu ^{\prime}=2/3$ . So the equation of the tangent line is $y=2/3(x-0)+1$ . The equation and its tangent line are graphed in Figure 2.6.3.

^†^†margin: Figure 2.6.3: A graph of the equation

y^{3}+x^{2}y^{4}=1+2x

along with its tangent line at the point

(0,1)

. Λ

Notice how our curve looks much different than other functions we have worked with up to this point. Such curves are important in many areas of mathematics, so developing tools to deal with them is also important.

Example 2.6.4 Using Implicit Differentiation

Given the implicitly defined curve $\sin(x^{2}y^{2})+y^{3}=x+y$ , find $y\mkern 1.35mu ^{\prime}$ .

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

	$\displaystyle\frac{\operatorname{d}\!}{\operatorname{d}\!x}\left(\sin(x^{2}y^{% 2})\right)$	$\displaystyle=\cos(x^{2}y^{2})\cdot\frac{\operatorname{d}\!}{\operatorname{d}% \!x}(x^{2}y^{2})$
		$\displaystyle=\cos(x^{2}y^{2})\cdot\left(x^{2}\frac{\operatorname{d}\!}{% \operatorname{d}\!x}(y^{2})+\frac{\operatorname{d}\!}{\operatorname{d}\!x}(x^{% 2})\cdot y^{2}\right)$
		$\displaystyle=\cos(x^{2}y^{2})\cdot(x^{2}\cdot 2yy\mkern 1.35mu ^{\prime}+2xy^% {2})$
		$\displaystyle=2(x^{2}yy\mkern 1.35mu ^{\prime}+xy^{2})\cos(x^{2}y^{2}).$

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

2(x^{2}yy\mkern 1.35mu ^{\prime}+xy^{2})\cos(x^{2}y^{2})+3y^{2}y\mkern 1.35mu % ^{\prime}=1+y\mkern 1.35mu ^{\prime}.

^†^†margin: (a) (b) Figure 2.6.4: A graph of the equation

\sin(x^{2}y^{2})+y^{3}=x+y

and certain tangent lines. Λ

We now have to be careful to properly solve for $y\mkern 1.35mu ^{\prime}$ , particularly because of the product on the left. It is best to multiply out the product. Doing this, we get

2x^{2}y\cos(x^{2}y^{2})y\mkern 1.35mu ^{\prime}+2xy^{2}\cos(x^{2}y^{2})+3y^{2}% y\mkern 1.35mu ^{\prime}=1+y\mkern 1.35mu ^{\prime}.

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

2x^{2}y\cos(x^{2}y^{2})y\mkern 1.35mu ^{\prime}+3y^{2}y\mkern 1.35mu ^{\prime}% -y\mkern 1.35mu ^{\prime}=1-2xy^{2}\cos(x^{2}y^{2}).

Then we can solve for $y\mkern 1.35mu ^{\prime}$ to get

y\mkern 1.35mu ^{\prime}=\frac{1-2xy^{2}\cos(x^{2}y^{2})}{2x^{2}y\cos(x^{2}y^{% 2})+3y^{2}-1}.

A graph of this implicit equation is given in Figure 2.6.4(a). 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 $y\mkern 1.35mu ^{\prime}$ .
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 2.6.4(b).

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 2.6.5 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 $(\frac{1}{2},\frac{\sqrt{3}}{2})$ .

SolutionTaking derivatives, we get $2x+2yy\mkern 1.35mu ^{\prime}=0$ . Solving for $y\mkern 1.35mu ^{\prime}$ gives:

\displaystyle y\mkern 1.35mu ^{\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\mkern 1.35mu ^{\prime}=\frac{-1/2}{\sqrt{3}/2}=\frac{-1}{\sqrt{3}}\approx-0.% 577.

In Figure 2.6.5, we see a graph of the circle and its tangent line at the point $(1/2,\sqrt{3}/2)$ 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.)

^†^†margin: Figure 2.6.5: The unit circle with its tangent line at

(1/2,\sqrt{3}/2)

. Λ

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.

Implicit Differentiation and the Second Derivative

We can use implicit differentiation to find higher order derivatives. In theory, this is simple: first find $\frac{\operatorname{d}\!y}{\operatorname{d}\!x}$ , 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 2.6.6 Finding the second derivative

Given $x^{2}+y^{2}=1$ , find $\displaystyle\frac{\operatorname{d}\!^{2}y}{\operatorname{d}\!x^{2}}=y\mkern 1% .35mu ^{\prime\prime}$ .

SolutionWe found that $y\mkern 1.35mu ^{\prime}=\frac{\operatorname{d}\!y}{\operatorname{d}\!x}=-x/y$ in Example 2.6.5. To find $y\mkern 1.35mu ^{\prime\prime}$ , we apply implicit differentiation to $y\mkern 1.35mu ^{\prime}$ .

$\displaystyle y\mkern 1.35mu ^{\prime\prime}$	$\displaystyle=\frac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}y\mkern 1.% 35mu ^{\prime}\bigr{)}$
	$\displaystyle=\frac{\operatorname{d}\!}{\operatorname{d}\!x}\left(-\frac{x}{y}\right)$	now use the Quotient Rule
	$\displaystyle=-\frac{y(1)-x(y\mkern 1.35mu ^{\prime})}{y^{2}}$	replace $y\mkern 1.35mu ^{\prime}$ with $-x/y$
	$\displaystyle=-\frac{y-x(-x/y)}{y^{2}}$
	$\displaystyle=-\frac{y^{2}+x^{2}}{y^{3}},$	since we were given $x^{2}+y^{2}=1$
	$\displaystyle=-\frac{1}{y^{3}}.$

We can see that $y\mkern 1.35mu ^{\prime\prime}>0$ when $y<0$ and $y\mkern 1.35mu ^{\prime\prime}<0$ when $y>0$ . In Section 3.4, we will see how this relates to the shape of the graph.

Implicit differentiation proves to be useful as it allows us to find the instantaneous rates of change of a variety of functions. We close with a small gallery of “interesting” and “famous” curves along with the implicit equations that produce them.


Astroid $x^{2/3}+y^{2/3}=1$	Fattened circle $x^{2n}+y^{2n}=1$	Elliptic curve $y^{2}=x^{3}+ax+b$

Cassini ovals $((x-a)^{2}+y^{2})((x+a)^{2}+y^{2})=b^{4}$	Devil’s curve $y^{2}(y^{2}-a^{2})=x^{2}(x^{2}-b^{2})$	Folium of Descartes $x^{3}+y^{3}=3axy$

The astroid and folium of Descartes also appear in Section 10.2. Other important implicit curves are the conic sections, which appear in Section 10.0. Also, the lemniscate, cardioid, and limaçon seen in Section 10.4 have particularly nice implicit representations.

In this chapter we have defined the derivative, given rules to facilitate its computation, and given the derivatives of a number of standard functions. We restate the most important of these in the following theorem, intended to be a reference for further work.

Theorem 2.6.1 Glossary of Derivatives of Elementary Functions

Let $u$ and $v$ be differentiable functions, and let $c$ and $n$ be real numbers, $n\neq 0$ .
1. $\frac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}cu\bigr{)}=cu^{\prime}$ 2. $\frac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}u\pm v\bigr{)}=u^{\prime% }\pm v^{\prime}$ 3. $\frac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}u\cdot v\bigr{)}=uv^{% \prime}+u^{\prime}v$ 4. $\frac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}\frac{u}{v}\bigr{)}=% \frac{u^{\prime}v-uv^{\prime}}{v^{2}}$ 5. $\frac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}u(v)\bigr{)}=u^{\prime}(% v)v^{\prime}$ 6. $\frac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}x^{n}\bigr{)}=nx^{n-1}$ 7. $\frac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}c\bigr{)}=0$ 8. $\frac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}x\bigr{)}=1$ 9. $\frac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}\ln x\bigr{)}=\frac{1}{x}$ 10. $\frac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}e^{x}\bigr{)}=e^{x}$ 11. $\frac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}\sin x\bigr{)}=\cos x$ 12. $\frac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}\cos x\bigr{)}=-\sin x$ 13. $\frac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}\tan x\bigr{)}=\sec^{2}x$ 14. $\frac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}\cot x\bigr{)}=-\csc^{2}x$ 15. $\frac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}\sec x\bigr{)}=\sec x\tan x$ 16. $\frac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}\csc x\bigr{)}=-\csc x\cot x$

Exercises 2.6

Terms and Concepts

1.

In your own words, explain the difference between implicit functions and explicit functions.
2.

Implicit differentiation is based on what other differentiation rule?
3.

T/F: Implicit differentiation can be used to find the derivative of $y=\sqrt{x}$ .
4.

T/F: Implicit differentiation can be used to find the derivative of $y=x^{3/4}$ .

Problems

In Exercises 5–22., find $\displaystyle\frac{\operatorname{d}\!y}{\operatorname{d}\!x}$ using implicit differentiation.

5.

$x^{4}+y^{2}+y=7$
6.

$x^{2/5}+y^{2/5}=1$
7.

$\cos x+\sin y=1$
8.

$\displaystyle\frac{x}{y}=10$
9.

$\displaystyle\frac{y}{x}=10$
10.

$x^{2}e^{2}+e^{y}=5$
11.

$x^{2}\tan y=50$
12.

$(3x^{2}+2y^{3})^{4}=2$
13.

$(y^{2}+2y-x)^{2}=200$
14.

$\displaystyle\frac{x^{2}+y}{x+y^{2}}=17$
15.

$\displaystyle\frac{\sin(x)+y}{\cos(y)+x}=1$
16.

$\ln(x^{2}+y^{2})=e$
17.

$\ln(x^{2}+xy+y^{2})=1$
18.

$xe^{x}=ye^{y}$
19.

$y\sin(x^{3})=x\sin(y^{3})$
20.

$\sqrt{xy}=1+x^{2}y$
21.

$x^{2}y-y^{2}x=1$
22.

$x^{2}+y^{2}+xy=7$

In Exercises 23–28., find the equation of the tangent line to the graph of the implicitly defined function at the indicated points. As a visual aid, each function is graphed.

23.
$x^{2/5}+y^{2/5}=1$ (a) At $(1,0)$ . (b) At $(0.1,0.281)$ (which does not exactly lie on the curve, but is very close).
24.
$x^{4}+y^{4}=1$ (a) At $(1,0)$ . (b) At $(\sqrt{0.6},\sqrt{0.8})$ . (c) At $(0,1)$ .
25.
$(x^{2}+y^{2}-4)^{3}=108y^{2}$ (a) At $(0,4)$ . (b) At $(2,-\sqrt[4]{108})$ .
26.
$(x^{2}+y^{2}+x)^{2}=x^{2}+y^{2}$ (a) At $(0,1)$ . (b) At $\displaystyle\left(-\frac{3}{4},\frac{3\sqrt{3}}{4}\right)$ .
27.
$(x-2)^{2}+(y-3)^{2}=9$ (a) At $\displaystyle\left(\frac{7}{2},\frac{6+3\sqrt{3}}{2}\right)$ . (b) At $\displaystyle\left(\frac{4+3\sqrt{3}}{2},\frac{3}{2}\right)$ .
28.
$x^{2}+2xy-y^{2}+x=2$ (a) At $(-2,0)$ . (b) At $(1,2)$ .

In Exercises 29–32., an implicitly defined function is given. Find $\displaystyle\frac{\operatorname{d}\!^{2}y}{\operatorname{d}\!x^{2}}$ . Note: these are the same problems used in Exercises 5.–8..

29.

$x^{4}+y^{2}+y=7$
30.

$x^{2/5}+y^{2/5}=1$
31.

$\cos x+\sin y=1$
32.

$\displaystyle\frac{x}{y}=10$

33.
Show that $\displaystyle\frac{\operatorname{d}\!y}{\operatorname{d}\!x}$ is the same for each of the following implicitly defined functions. (a) $xy=1$ (b) $x^{2}y^{2}=1$ (c) $\sin(xy)=1$ (d) $\ln(xy)=1$