implicit differentiation

Differentiation, derivatives and Taylor approximations: Differentiating implicit functions

Implicit differentiation

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.

Slope of a circle at a point 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 \(-\frac{x}{y}\biggl|_{(-3,4)}=-\frac{-3}{4}=\frac{3}{4}\).

Tangent line of a curve at a point 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

\sqrt{1 - a^{2}} y + a x = 1 .

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=x#

We differentiate the left- and right-hand side of the equation \[-y^3=x\] with respect to #x# and consider #y# as a differentiable function in #x#: \[ \frac{\dd}{\dd x}\!\left(-y^3\right) = \frac{\dd}{\dd x}\!\left(x^{\phantom{2}}\!\!\!\right)\] Applying the chain rule then yields: \[-3y^2\cdot\dfrac{\dd y}{\dd x} = 1\] So: \[\dfrac{\dd y}{\dd x}=-{{1}\over{3y^2}}\]

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.