Calculus I 02.05 Implicit Differentiation

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Implicit Differentiation

• Distinguish between functions written in implicit and explicit form.
• Use implicit differentiation to find the derivative.

Implicit and Explicit Functions

So far in the discussion functions were expressed in explicit form where \(y\) is equal to some function. For example, \(y=3x^2-5\) has the \(y\) variable explicitly written as a function for \(x\). With the implicit form the \(y\) variable is not isolated. For example, \(xy=1\) is a function where \(y\) is in the function. To find \(dy/dx\) for an implicit function rewrite it as an explicit function and differentiate.

Implicit Form	Explicit Form	Derivative
\(xy=1\)	$$y=\frac{1}{x}=x^{-1}$$	$$\frac{dy}{dx}=-x^{-2}=-\frac{1}{x^2}$$

This strategy fails when \(y\) cannot be expressed as a function for \(x\). For example, how can

$$x^2-2y^3+4xy=2$$

be differentiated? By using implicit differentiation.

Differentiation always takes place with respect to \(x\). When differentiating terms involving \(y\) the Chain Rule must be applied. The assumption is that \(y\) is defined implicitly as a differentiable function for \(x\).

Example 2.5.1 Differentiating with Respect to \(x\)

Implicit Differentiation

Guidelines for Implicit Differentiation
1.	Differentiate both sides with respect to \(x\).
2.	Move all terms involving \(dy/dx\) to the left and all other terms to the right.
3.	Factor \(dy/dx\) from the left side.
4.	Solve for \(dy/dx\).

Example 2.5.2 Implicit Differentiation Guidelines

Figure 2.5.1

Find \(dy/dx\) given that \(y^3+y^2-5y-x^2=-1.\) Solution 1. Differentiate both sides with respect to \(x\). $$\frac{d}{dx}[y^3+y^2-5y-x^2] $$ $$=\frac{d}{dx}[-4]$$ $$\frac{d}{dx}[y^3]+\frac{d}{dx}[y^2]-\frac{d}{dx}[5y]-\frac{d}{dx}[x^2]$$ $$=\frac{d}{dx}[-4]$$ $$3y^2\frac{dy}{dx}+2y\frac{dy}{dx}-5\frac{dy}{dx}-2x$$ \(=0\) 2. Collect the \(dy/dx\) terms on the left and move all other terms to the right. $$3y^2\frac{dy}{dx}+2y\frac{dy}{dx}-5\frac{dy}{dx}=2x$$ 3. Factor \(dy/dx\) out from the left side. $$\frac{dy}{dx}(3y^2+2y-5)=2x$$ 4. Solve for \(dy/dx\) by dividing by \((3y^2+2y-5)\). $$\frac{dy}{dx}=\frac{2x}{3y^2+2y-5}$$ Even when \(y\) is not a function for \(x\) the derivative produces a formula for the tangent line slope at several points, as shown in Figure 2.5.1.

Some graph segments can be represented by differentiable functions. Figure 2.5.2

Solving for \(dy/dx\) is meaningless in an equation with no solution points. For example, \(x^2+y^2=-4\) has no solution points. If a graph segment is represented by a differentiable function, then \(dy/dx\) will have meaning as the slope at each point on the segment. A function is not differentiable at points with vertical tangents and points where the function is not continuous. Example 2.5.3 Graphs and Differentiable Functions If possible, represent \(y\) as a differentiable function for \(x\). a. \(x^2+y^2=0\)    b. \(x^2+y^2=1\)   c. \(x+y^2=1\) Solution a. The graph for this equation is a single point. Therefore, it does not define \(y\) as a differentiable function for \(x\), as shown in Figure 2.5.2(a). b. The graph for this equation is the unit circle centered at \((0,0)\). The upper semicircle is given by the differentiable function \(y=\sqrt{1-x^2},\:\:\:\:-1<x<1.\) and the lower semicircle is given by the differentiable function \(y=-\sqrt{1-x^2},\:\:\:\:-1<x<1.\) At the point \((-1,0)\) and \((1,0)\), the slope for the graph is undefined, as shown in Figure 2.5.2(b). c. The parabola's upper half is given by the differentiable function \(y=\sqrt{1-x},\:\:\:\:x<1\) and the lower half is given by the differentiable function \(y=-\sqrt{1-x},\:\:\:\:x<1.\) At the point \((1,0)\), the graph's slope is undefined, as shown in Figure 2.5.2(c).

Example 2.5.4 Implicitly Finding the Slope for a Graph

Figure 2.5.3

Determine the tangent line's slope to the graph for \(x^2+4y^2=4\) at the point \((\sqrt{2},-1\sqrt{2})\), as shown in Figure 2.5.3. Solution \(x^2+4y^2\) \(=4\)     The original equation. $$2x+8y\frac{dy}{dx}$$ \(=0\)     Differentiable with respect to \(x\). $$\frac{dy}{dx}$$ $$=\frac{-2}{8y}$$     Solve for \(dy/dx\). $$=\frac{-x}{4y}$$     Simplify. At \((\sqrt{2}, -1\sqrt{2})\), the slope is $$\frac{dy}{dx}$$ $$=\frac{-\sqrt{2}}{-4/\sqrt{2}}=\frac{1}{2}.$$     Evaluate \(dy/dx\) when \(x=\sqrt{2}\) and \(y=-1/\sqrt{2}\).

Example 2.5.5 Implicitly Finding the Slope for a Graph

Leminiscale Figure 2.5.4

Determine the slope to the graph for \(3(x^2+y^2)^2=100xy) at the point \((3,1)\), as shown in Figure 2.5.4. Solution At the point \((3,1)\), the slope for the graph is $$\frac{dy}{dx}=\frac{25(1)-3(3)(3^2+1^2)}{-25(3)+3(1)(3^2+1^2)}=\frac{25-90}{-75+3-}=frac{-65}{-45}=\frac{13}{9}$$ as shown in Figure 2.5.4. This graph is called a lemniscate graph.

Example 2.5.6 Determining a Differentiable Function

The derivative is \(\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}\). Figure 2.5.5

Find \(dy/dx\) implicitly for the equation \(\sin y= x.\) Then find the largest interval for the form \(-a < y < a\) on which \(y\) is a differentiable function for \(x\), as shown in Figure 2.5.5. Solution $$\frac{d}{dx}[\sin y]$$ $$=\frac{d}{dx}[x]$$ $$\cos y\frac{dy}{dx}$$ \(=1\) $$\frac{dy}{dx}$$ $$=\frac{1}{\cos y}$$ The largest interval bout the origin for which \(y\) is a differentiable function for \(x\) is \(-\pi /2 < y < \pi/2\). Note that \(\cos y\) is positive for all \(y\) in this interval and is 0 at the endpoints. When restricting \(y\) to the interval \(-\pi/2 < y < \pi /2\) \(dy/dx\) can be written explicitly as a function for \(x\). Use the identity \(\cos y\) \(=\sqrt{1 - \sin^2 y}\) $$= \sqrt{1-x^2},\:\:\: -\frac{\pi}{2} < y < \frac{\pi}{2}$$ and conclude that $$\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$$

Example 2.5.7 Finding the Second Derivative Implicitly

Given

\(x^2+y^2=25,\)

find

$$\frac{d^2y}{dx^2}.$$

Solution Differentiating each term with respect to \(x\) produces

Differentiating a second time with respect to \(x\) yields

$$\frac{d^2y}{dx^2}$$	$$=-\frac{(y)(1)-(x)(dy/dx)}{y^2}$$	Quotient Rule
	$$=-\frac{y-(x)( \color{red}{-x/y})}{y^2}$$	Substitute \(-x/y\) for \(dy/dx\).
	$$=-\frac{y^2+x^2}{y^2}$$	Simplify.
	$$=-\frac{\color{red}{25}}{y^3}.$$	Substitute 25 for \(x^2+y^2.\)

Example 2.5.8 Finding a Tangent Line to a Graph

The kappa curve Figure 2.5.6

Find the tangent line to the graph for \(x^2(x^2+y^2)=y^2\) at the point \((\sqrt{2}/2,\sqrt{2}/2)\), as shown in Figure 2.5.6. Solution Rewriting and differentiating implicitly produces At the point \((\sqrt{2}/2,\sqrt{2}/2)\), the slope is $$\frac{dy}{dx}=\frac{(\sqrt{2}/2)[2(1/2)+(1/2)]}{(\sqrt{2}/2)[1-(1/2)]}=\frac{3/2}{1/2}=3$$ and the equation for the tangent line at this point is

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Parent Article: Calculus I 02 Differentiation