Lesson 13: Implicit Differentiation
Implicit Differentiation
Definition: A function is in explicit form if it is written in the form \(y = f(x)\), that is, \(y\) is isolated on one side of the equal sign and everything else that does not involve \(y\) is on the other side.
Example 1
The function \(y = 2x + 7\) is in explicit form. Find \(y' = \frac{dy}{dx}\).
Solution:
Taking the derivative of both sides with respect to \(x\):
\[\frac{dy}{dx} = \frac{d}{dx}(2x + 7) = 2\]Therefore, \(y' = 2\).
Definition: A function is in implicit form if it is not in explicit form.
Important Note: Because \(y\) is a function of \(x\) (i.e. \(y = f(x)\)), when we take the derivative of \(y\) with respect to \(x\), we have that \(y' = \frac{dy}{dx}\), not 1. Think of \(y\) as \(y(x)\).
Example 1 (revisited)
The function \(y - 2x = 7\) is in implicit form. Find \(y' = \frac{dy}{dx}\).
Solution:
Taking the derivative of both sides with respect to \(x\):
\[\frac{d}{dx}(y - 2x) = \frac{d}{dx}(7)\]Applying the derivative to each term:
\[\frac{dy}{dx} + \frac{d}{dx}(-2x) = \frac{d}{dx}(7)\implies\frac{dy}{dx} - 2 = 0\]Solving for \(\frac{dy}{dx}\):
\[y' = \frac{dy}{dx} = 2\]Note: We call the technique used in the previous example implicit differentiation. The key is to take the derivative with respect to \(x\) of both sides of the equation. This technique is especially useful when we cannot put an equation into explicit form.
Example 2
Let \(3y^2 + 2xy = 4x + 3\). Find \(y'\).
Solution:
Taking the derivative of both sides with respect to \(x\):
\[\frac{d}{dx}(3y^2 + 2xy) = \frac{d}{dx}(4x + 3)\implies \frac{d}{dx}(3y^2)+\frac{d}{dx}(2xy)=\frac{d}{dx}(4x)+\frac{d}{dx}(3)\]We use the chain rule to solve for \(\frac{d}{dx}(3y^2)\), since \(y\) is a function of \(x\), yielding \(6yy'\). To solve for \(\frac{d}{dx}(2xy)\), we apply the product rule, yielding \(2y + 2xy'\). So, we have that
\[6yy' + (2y + 2xy') = 4 + 0.\]Collecting terms with \(y'\):
\[6yy' + 2xy' = 4 - 2y\]Factoring out \(y'\):
\[y'(6y + 2x) = 4 - 2y\]Solving for \(y'\):
\[y' = \frac{4 - 2y}{6y + 2x}\]Example 3
Let \(3\sin\left(\frac{y}{x}\right) = 7x\). Find \(y'\).
Solution:
Taking the derivative of both sides with respect to \(x\):
\[\frac{d}{dx}\left(3\sin\left(\frac{y}{x}\right)\right) = \frac{d}{dx}(7x)\]Applying the chain rule and quotient rule on the left side of the equation:
\[3\cos\left(\frac{y}{x}\right) \cdot \frac{d}{dx}\left(\frac{y}{x}\right) = 7\]Using the quotient rule to solve for \(\frac{d}{dx}(\frac{y}{x})\):
\[3\cos\left(\frac{y}{x}\right) \cdot \frac{xy' - y \cdot 1}{x^2} = 7 \implies 3\cos\left(\frac{y}{x}\right) \cdot \frac{xy' - y}{x^2} = 7\]Solving for \(y'\):
\[\begin{align}\frac{xy' - y}{x^2} = \frac{7}{3\cos\left(\frac{y}{x}\right)}&\implies xy' - y = \frac{7x^2}{3\cos\left(\frac{y}{x}\right)}\\ &\implies xy' = \frac{7x^2+3y\cos\left(\frac{y}{x}\right)}{3\cos\left(\frac{y}{x}\right)}\\ &\implies y' = \frac{7x^2 + 3y\cos\left(\frac{y}{x}\right)}{3x\cos\left(\frac{y}{x}\right)}\end{align}\]Example 4
Let \(e^{6xy} = 7x\). Find \(y'\).
Solution:
Taking the derivative of both sides with respect to \(x\) and applying the chain rule and product rule:
\[\frac{d}{dx}(e^{6xy}) = \frac{d}{dx}(7x) \implies e^{6xy} \cdot \frac{d}{dx}(6xy) = 7 \implies e^{6xy} \cdot (6y + 6xy') = 7\]Solving for \(y'\):
\[6y + 6xy' = \frac{7}{e^{6xy}} \implies 6xy' = \frac{7}{e^{6xy}} - 6y \implies y' = \frac{7 - 6ye^{6xy}}{6xe^{6xy}}\]Example 5
Let \(8\cos(x)\sin(y) = 5\). Find \(y'\).
Solution:
Taking the derivative of both sides with respect to \(x\):
\[\frac{d}{dx}(8\cos(x)\sin(y)) = \frac{d}{dx}(5)\]Applying the product rule on the left side:
\[8\left(-\sin(x)\sin(y) + \cos(x)\cos(y)y'\right) = 0\implies -8\sin(x)\sin(y) + 8\cos(x)\cos(y)y' = 0\]Solving for \(y'\):
\[8\cos(x)\cos(y)y' = 8\sin(x)\sin(y)\implies y' = \frac{8\sin(x)\sin(y)}{8\cos(x)\cos(y)}\implies y' = \tan(x)\tan(y)\]Example 6
Find the slope of the tangent line to the graph of \(2x^4 = 3y^2 - y\) at the point \((1, 1)\).
Solution:
First, we find \(y'\) using implicit differentiation:
\[\frac{d}{dx}(2x^4) = \frac{d}{dx}(3y^2 - y)\implies 8x^3 = 6yy' - y'\implies 8x^3 = y'(6y - 1)\implies y' = \frac{8x^3}{6y - 1}\]Now we evaluate at the point \((1, 1)\):
\[y'\bigg|_{(1,1)} = \frac{8\cdot 1^3}{6\cdot 1 - 1} = \frac{8}{5}\]Therefore, the slope of the tangent line at \((1, 1)\) is \(\frac{8}{5}\).
Example 7
Suppose \(2\sin(6 x + 7y) = 3xy\). Find \(\frac{dy}{dx}\).
Solution:
Taking the derivative of both sides with respect to \(x\):
\[\frac{d}{dx}\left(2\sin(6 x + 7y)\right) = \frac{d}{dx}(3xy)\]Applying the chain rule on the left (treating \(6 x + 7y\) as the inner function) and the product rule on the right:
\[2\cos(6 x + 7y) \cdot \frac{d}{dx}(6 x + 7y) = 3y + 3xy'\implies 2\cos(6 x + 7y) \cdot (6 + 7y') = 3y + 3xy'\]Expanding:
\[12\cos(6 x + 7y) + 14y'\cos(6 x + 7y) = 3y + 3xy'\]Collecting terms with \(y'\):
\[14y'\cos(6 x + 7y) - 3xy' = 3y - 12\cos(6 x + 7y)\]Factoring out \(y'\):
\[y'(14\cos(6 x + 7y) - 3x) = 3y - 12\cos(6 x + 7y)\]Solving for \(y'\):
\[y' = \frac{3y - 12\cos(6 x + 7y)}{14\cos(6 x + 7y) - 3x}\]Example 8
Suppose \(\frac{5}{x} + \frac{1}{6y} = 7\). Find \(\frac{dy}{dx}\bigg|_{(1, \frac{1}{12})}\).
Solution:
First, we rewrite the equation:
\[5x^{-1} + \frac{1}{6}y^{-1} = 7\]Taking the derivative of both sides with respect to \(x\):
\[\frac{d}{dx}\left(5x^{-1} + \frac{1}{6}y^{-1}\right) = \frac{d}{dx}(7)\implies -5x^{-2} - \frac{1}{6}y^{-2}y' = 0\]Solving for \(y'\):
\[-\frac{1}{6}y^{-2}y' = 5x^{-2}\implies y' = -\frac{5x^{-2}}{\frac{1}{6}y^{-2}} = -5x^{-2} \cdot 6y^{2} = -\frac{30y^2}{x^2}\]Evaluating at the point \(\left(1, \frac{1}{12}\right)\):
\[\frac{dy}{dx}\bigg|_{(1, \frac{1}{12})} = -\frac{30\left(\frac{1}{12}\right)^2}{1^2} = -\frac{30 \cdot \frac{1}{144}}{1} = -\frac{30}{144} = -\frac{5}{24}\]