Economics at Zero Marginal Cost: Differentiation of Implicit Function

Explicit and Implicit Function

A function given in the form of $y=f(x)$ , say, $y=f(x)=3x^4$ is called an explicit function, because the variable $y$ is explicitly expressed as a function of $x$ .

A function in which the dependent variable is not isolated on one side of the equation is known as an implicit function. For example, the equation $x^2+xy-y^2=1$ represents an implicit function. Implicit functions are usually given in terms of both dependent and independent variables.

Exercise:

Find $\frac{dy}{dx}$

1. $x^2y^3+3xy+y+2=0\\$

Solution:

Given:

$x^2y^3+2xy+y+2=0\\$

Differentiating both sides with respect to $x$ .

$\\\frac{d}{dy}(x^2y^3+2xy+y+2)=\frac{d}{dx}0\\ \frac{d}{dx}x^2y^3+\frac{d}{dx}2xy +\frac{d}{dx}y+\frac{d}{dy}2=0\\ (x^2\frac{d}{dx}y^3+y^3\frac{d}{dx}x^2)+2\frac{d}{dx}xy+\frac{dy}{dx}+0=0\\ x^23y^{3-1}\frac{d}{dx}y+y^32x^{2-1}+2\left ( x\frac{d}{dx}y+y\frac{d}{dx} \right )+\frac{dy}{dx}=0\\ 3x^2y^2.\frac{dy}{dx}+2xy^3+2\left ( x.\frac{dy}{dx}+y.1 \right )+\frac{dy}{dx}=0\\ 3x^2y^2.\frac{dy}{dx}+2xy^3+2\left ( x.\frac{dy}{dx}+y \right )+\frac{dy}{dx}=0\\ 3x^2y^2.\frac{dy}{dx}+2xy^3+2x.\frac{dy}{dx}+2y+\frac{dy}{dx}=0\\ 3x^2y^2.\frac{dy}{dx}+2x.\frac{dy}{dx}+\frac{dy}{dx}=2xy^3-2y\\ (3x^2y^2+2x+1)\frac{dy}{dx}=-(2xy^3+2y)\\ \frac{dy}{dx}=-\frac{2xy^3+2y}{3x^2y^2+2x+1}$

2. $x^2- 7xy+y^2=10$

Solution:

Given:

$x^2- 7xy+y^2=10$

Differentiating both sides with respect to $x$ .

$\\\frac{d}{dy}(x^2- 7xy+y^2)=\frac{d}{dx}10\\ \frac{d}{dx}x^2-\frac{d}{dx}7xy+\frac{d}{dx}y^2=0\\ 2x^{2-1}-7\frac{d}{dx}xy+2y^{2-1}.\frac{d}{dx}y=0\\ 2x-7\left ( x\frac{d}{dx}y+y\frac{d}{dx}x \right )+2y.\frac{dy}{dx}=0\\ 2x-7\left ( x.\frac{dy}{dx}+y.1 \right )+2y.\frac{dy}{dx}=0\\ 2x-7\left ( x\frac{dy}{dx}+y \right )+2y.\frac{dy}{dx}0\\ 2x-7x.\frac{dy}{dx}-7y+2y.\frac{dy}{dx}+0\\ -7x.\frac{dy}{dx}+2y.\frac{dy}{dx}=7y-2x\\ (2y-7x)\frac{dy}{dx}=7y-2x\\ \frac{dy}{dx}=\frac{7y-2x}{2y-7x}$

3. $xy-(x+y+1)^6=0$

Solution:

Given:

$xy-(x+y+1)^6=0$

Differentiating both sides with respect to $x$ .

$\\\frac{d}{dx}[xy-(x+y+1)^6]=\frac{d}{dx}0\\ \frac{d}{dx}xy-\frac{d}{dx}(x+y+1)^6=0\\ x\frac{d}{dx}y+y\frac{d}{dx}x-6(x+y+1)^{6-1}.\frac{d}{dx}(x+y+1)=0\\ x\frac{dy}{dx}+y.1-6(x+y+1)^5\left ( \frac{d}{dx}x+\frac{d}{dx}y+\frac{d}{dx}1 \right )=0\\ x\frac{dy}{dx}+y-6(x+y+1)^5\left ( 1+\frac{dy}{dx}+0 \right )=0\\ x\frac{dy}{dx}+y-6(x+y+1)^5\left ( 1+\frac{dy}{dx} \right )=0\\ x\frac{dy}{dx}+y-6(x+y+1)^5-6(x+y+1)^5\frac{dy}{dx}=0\\ x\frac{dy}{dx}-6(x+y+1)^5\frac{dy}{dx}=6(x+y+1)^5-y\\ \left [x-6(x+y+1)^5 \right ]\frac{dy}{dx}=6(x+y+1)^5-y\\ \frac{dy}{dx}=\frac{6(x+y+1)^5-y}{x-6(x+y+1)^5}$

4. $x^{1/2}+y^{1/2}=a^{1/2}$

Solution:

Given:

$x^{1/2}+y^{1/2}=a^{1/2}$

Differentiating both sides with respect to $x$ .

$\\\frac{d}{dx}(x^{1/2}+y^{1/2})=\frac{d}{dx}a^{1/2}\\ \frac{d}{dx}x^{1/2}+\frac{d}{dx}y^{1/2}=0\\ \frac{1}{2}x^{\frac{1}{2}-1}+\frac{1}{2}y^{\frac{1}{2}-1}.\frac{d}{dx}y=0\\ \frac{1}{2}x^{-\frac{1}{2}}+\frac{1}{2}y^{-\frac{1}{2}}.\frac{dy}{dx}=0\\ \frac{1}{2}y^{-\frac{1}{2}}.\frac{dy}{dx}=-\frac{1}{2}x^{-\frac{1}{2}}\\ \frac{dy}{dx}=\frac{-\frac{1}{2}x^{-\frac{1}{2}}}{\frac{1}{2}y^{-\frac{1}{2}}}\\ \frac{dy}{dx}=\frac{x^{-\frac{1}{2}}}{y^{-\frac{1}{2}}}\\ \frac{dy}{dx}=-\frac{y^{\frac{1}{2}}}{x^{\frac{1}{2}}}\\ \frac{dy}{dx}=-\frac{\sqrt{y}}{\sqrt{x}}\\$

Economics at Zero Marginal Cost

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Wednesday, June 10, 2020

Differentiation of Implicit Function

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