implicit differentiation

Differentiation, derivatives and Taylor approximations: Differentiating implicit functions

Implicit differentiation

We have already noted that quantities are often related to each other through an implicit relationship. For example, you can write a linear relationship between $x$ and $y$ as the equation

$a\,x+b\,y+c=0$ for certain constants $a$ , $b$ and $c$ . If $b\neq 0$ , then you can isolate $y$

$y=-\frac{a}{b}x-\frac{c}{b}$ and then consider $y$ as a function of $x$ . For the derivative of $y$ as a function of $x$ you get:

$\frac{\dd y}{\dd x}=-\frac{a}{b}$

But we could also have determined this derivative by applying the differential operator $\frac{\dd }{\dd x}$ directly to the implicit relationship between $x$ and $y$ and using calculation rules for differentiation:

$\begin{aligned}\frac{\dd }{\dd x}(a\,x+b\,y+c)&=\frac{\dd }{\dd x}(a\,x)+\frac{\dd }{\dd x}(b\,y)+ \frac{\dd }{\dd x}(c)\\&\phantom{abcpqrxyz}\blue{\text{sum rule for differentiation}}\\[0.25cm] &= a\frac{\dd x}{\dd x}+ b\frac{\dd y}{\dd x}+0&\\&\phantom{abcpqrxyz}\blue{\text{constant factor rule, derivative of a constant}}\\[0.25cm] &= a+b\frac{\dd y}{\dd x}\end{aligned}$ Because of $a\,x+b\,y+c=0$ , the previous result must also be equal to zero. The derivative can then be isolated in $a+b\frac{\dd y}{\dd x}=0$ and you get the same result as before. The second approach is called the method of implicit differentiation.

Impliciete differentiation

If the graph of a function $f$ on an open interval $I$ of real numbers is part of the curve $F(x, y) = 0$ , then the value of the implicit derivative of $y$ at a point $c$ in $I$ coincides with $f'(c)$ .

The proof is based on the fact that all laws and rules used to calculate $\frac{\dd y}{\dd x}$ also apply to $f'(c)$ .

In the example of the unit circle, where $F(x, y) = x^2 + y^2 - 1$ , we differentiate the left and right sides with respect to $x$ (indicated by $\frac{\dd}{\dd x}$ ) and use the standard rules of differentiation:

$\begin{aligned} \ \frac{\dd}{\dd x}(x^2+y^2) &=\frac{\dd}{\dd x}(1)\\[0.2cm] 2x+2y\frac{\dd y}{\dd x} &= 0\\[0.2cm] y\frac{\dd y}{\dd x} &= -x\\[0.2cm] \frac{\dd y}{\dd x} &= -\dfrac{x}{y}\end{aligned}$

This technique is useful for determining the tangent line to the curve at any given point. Just like in the case of an explicit function, the tangent line at a point $(a, b)$ indicates the direction of movement along the curve from $(a, b)$ .

Suppose that $(a, b)$ is a point on the curve given by the equation $F(x, y) = 0$ . If the implicit derivative $\frac{\dd y}{\dd x}$ is defined at $a$ , then the tangent line to the curve at that point is given by the equation $y = y'(a)(x - a) + b$ , where $y'(a) = \frac{\dd y}{\dd x} (a)$ is the value at $a$ of the derivative of $y$ with respect to $x$ determined by $F(x, y) = 0$ .

To illustrate, we provide three examples of implicit differentiation.

We look at the circle with equation $x^2+y^2=25$ and ask ourselves what is the slope of the tangent line at the point $(-3,4)$ .

We consider $y$ as a function of $x$ in the given equation and apply the differential operator $\frac{\dd }{\dd x}$ to the left- and right-hand side of the equation; applying the rules of differentiation gives:

$\begin{aligned}\frac{\dd }{\dd x}(x^2)+\frac{\dd }{\dd x}(y^2)&=\frac{\dd }{\dd x}(25)\\[0.25cm] 2x+2y\frac{\dd y }{\dd x}&=0\\[0.25cm] \frac{\dd y }{\dd x} &= -\frac{x}{y}\end{aligned}$ The slope in $(-3,4)$ is therefore $-\frac{x}{y}\biggl|_{(-3,4)}=-\frac{-3}{4}=\frac{3}{4}$ .

We look at the curve with equation $x^3+xy+y^3=5$ and want to know the equation of the tangent line at the point $(2,-1)$ .

Substitution of $x=2$ and $y=-1$ in the equation of the curve shows that $(2,-1)$ is indeed a point on the curve. We consider $y$ as a function of $x$ in the given equation and apply the differential operator $\frac{\dd }{\dd x}$ left and right on the equation; application of the rules of differentiation gives:

$\begin{aligned}\frac{\dd }{\dd x}(x^3)+\frac{\dd }{\dd x}(xy)+\frac{\dd }{\dd x}(y^3)&=\frac{\dd }{\dd x}(5)\\[0.25cm] 3x^2+y+ x\frac{\dd y}{\dd x}+3y^2\frac{\dd y }{\dd x}&=0\\[0.25cm] 3x^2+y+ (x+3y^2)\frac{\dd y }{\dd x}&=0\end{aligned}$ Substitution of $x=2$ and $y=-1$ gives:

$\begin{aligned}3\times 2^2-1+\bigl(2+3\times(-1)^2\bigr)y'(2)&=0\\[0.25cm] 11+5y'(2)&=0\\[0.25cm]y'(2)&=-\frac{11}{5}\end{aligned}$ The equation of the tangent line at the point $(2,-1)$ is

$y=-\tfrac{11}{5}(x-2)-1\qquad\text{or}\qquad y=-\tfrac{11}{5}x+\tfrac{17}{5}$

Consider the point $(a, b) = (a, \sqrt{1-a^2})$ on the unit circle, whose equation is $x^2 + y^2 = 1$ . We have: $2x + 2y\frac{\dd y}{\dd x} = 0$ , or $y' = -\frac{x}{y}$ . Thus, $y'(a) = -\frac{a}{\sqrt{1-a^2}}$ . The equation for the tangent line becomes $y = -\frac{a}{\sqrt{1-a^2}}(x-a) + \sqrt{1-a^2}$ , or

1 - a^{2}

y + a x = 1 .

This can be verified by taking the equations for the curve and the line together and solving for $x$ after eliminating $y$ . It will be found that the quadratic equation in $x$ has only one solution $(x=a)$ . This confirms that the line is a tangent line to the curve at the point $(a, b)$ .

Look at more examples to illustrate the technique of implicit differentiation and its applications.

Determine the derivative $\dfrac{\dd y}{\dd x}$ in case $-y^3=-3x$

We differentiate the left- and right-hand side of the equation

$-y^3=-3x$ with respect to $x$ and consider $y$ as a differentiable function in $x$ :

$\frac{\dd}{\dd x}\!\left(-y^3\right) = \frac{\dd}{\dd x}\!\left(-3x^{\phantom{2}}\!\!\!\right)$ Applying the chain rule then yields:

$-3y^2\cdot\dfrac{\dd y}{\dd x} = -3$ So:

$\dfrac{\dd y}{\dd x}={{1}\over{y^2}}$

New example

A ladder is leaning against a wall and begins to slide down. On the ground, the ladder is positioned $x(t)$ metres away from the wall at time $t$ . The point of contact with the wall is at a height of $y(t)$ at time $t$ . The ladder is $5$ metres long, and at time $t=0$ , we have $x(0) = 3$ and $y(0) = 4$ . Note that $x(t)^2 + y(t)^2 = 25$ , and therefore, $y(t) = \sqrt{25 - x(t)^2}$ . Now, $x(t)$ is sliding away from the wall at a speed of, say, $1$ metre per unit of time:

$x'(t) = 1\tiny.$ Differentiating the equation

$x(t)^2 + y(t)^2 = 25$ with respect to $t$ , we get

$2 x(t)\cdot x'(t) + 2 y(t)\cdot y'(t) = 0$ At $t=0$ , this becomes

$2 x(0)\cdot x'(0) + 2 y(0)\cdot y'(0) = 6 + 8 y'(0) = 0$ so that $y'(0) = -\frac{3}{4}$ .

We can also solve

$y'(t) = - \frac{x(t)\cdot x'(t)}{y(t)} = -\frac{x(t)\cdot x'(t)}{\sqrt{25 - x(t)^2}} = -\frac{3 + t}{\sqrt{25 - (3+t)^2}}$ The last equality follows from $x'(t)=1$ and $x(0)=3$ , so $x(t)=3+t$ . Again, at time $t=0$ , this yields the result $y'(0) = -\frac{3}{4}$ metres per unit of time.