Economics at Zero Marginal Cost: Quotient Rule

The derivative of the quotient of two functions $\frac{f(x)}{g(x)}$ is $\frac{d}{dx}\frac{f(x)}{g(x)}=\frac{g(x)\frac{d}{dx}f(x)-f(x)\frac{d}{dx}g(x)}{g^2(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g^2(x)}$

Excercise:

Find $\frac{dy}{dx}$

$\\\1.\;y= \frac{(ax^2+b)}{(cx+d)}\\ \2.\;f(x)=\frac{x^{-3}}{x^2-5x+7}\\ \3\;y=\frac{x^3+3x+15}{x^2-2x+2}\\ \4\;y=\frac{3x^2-5x+6}{x^2}\\ \5\;y=\frac{\sqrt{x}+10}{\sqrt{x}-10}$

Excercise:

Find $\frac{dy}{dx}$ of:

1. If $y=\frac{(ax^2+b)}{(cx+d)}$

2. If $f(x)=\frac{x^{-3}}{x^2-5x+7}\\$

Solution:

Let $y=f(x)=[\frac{x^{-3}}{x^2-5x+7}]\\$

$\\\frac{dy}{dx}=\frac{d}{dx}y=\frac{d}{dx}\left [\frac{x^{-3}}{x^2-5x+7} \right ]\\ \frac{dy}{dx}=\frac{(x^2-5x+7)\frac{d}{dx}x^{-3}-x^{-3}\frac{d}{dx}(x^2-5x+7)}{(x^2-5x+7)^2}\\ \frac{dy}{dx}=\frac{(x^2-5x+7).-3x^{-3-1}-x^{-3}(\frac{d}{dx}x^2-\frac{d}{dx}5x+\frac{d}{dx}7)}{(x^2-5x+7)^2}\\ \frac{dy}{dx}=\frac{(x^2-5x+7).-3x^{-4}-x^{-3}(2x^{2-1}-5\frac{d}{dx}x+0)}{(x^2-5x+7)^2}\\ \frac{dy}{dx}=\frac{(-3x^{-2}+15x^{-3}-21x^{-4})-x^{-3}(2x-5.1)}{(x^2-5x+7)^2}\\ \frac{dy}{dx}=\frac{(-3x^{-2}+15x^{-3}-21x^{-4})-x^{-3}(2x-5)}{(x^2-5x+7)^2}\\ \frac{dy}{dx}=\frac{(-3x^{-2}+15x^{-3}-21x^{-4})-(2x^{-2}-5x^{-3})}{(x^2-5x+7)^2}\\ \frac{dy}{dx}=\frac{-3x^{-2}+15x^{-3}-21x^{-4}-2x^{-2}+5x^{-3}}{(x^2-5x+7)^2}\\ \frac{dy}{dx}=\frac{-5x^{-2}+20x^{-3}-21x^{-4}}{(x^2-5x+7)^2}$

3. If $y=\frac{x^3+3x+15}{x^2-2x+2}$

Solution:

$\\\frac{dy}{dx}=\frac{d}{dx}y=\frac{d}{dx}[\frac{x^3+3x+15}{x^2-2x+2}]\\ \frac{dy}{dx}=\frac{(x^2-2x+2)\frac{d}{dx}(x^3+3x+15)-(x^3+3x+15)\frac{d}{dx}(x^2-2x+2)}{x^2-2x+2^2}\\ \frac{dy}{dx}=\frac{(x^2-2x+2)\left ( \frac{d}{dx}x^3+\frac{d}{dx}3x+\frac{d}{dx}15\right )-(x^3+3x+15)\left ( \frac{d}{dx}x^2-\frac{d}{dx}2x+\frac{d}{dx}2\right )}{x^2-2x+2^2}\ \frac{dy}{dx}=\frac{(x^2-2x+2)\left ( 3x^{3-1}+3\frac{d}{dx}x+0\right )-(x^3+3x+15)(2x^{2-1}-2\frac{d}{dx}x+0)}{x^2-2x+2^2}\\ \frac{dy}{dx}=\frac{(x^2-2x+2)(3x^2+3.1)-(x^3+3x+15)(2x-2.1)}{x^2-2x+2^2}\\ \frac{dy}{dx}=\frac{(x^2-2x+2)(3x^2+3)-(x^3+3x+15)(2x-2)}{x^2-2x+2^2}\\ \frac{dy}{dx}=\frac{(3x^4+3x^2-6x^3-6x+6x^2+6)-(2x^4-2x^3+6x^2-6x+30x-30)}{x^2-2x+2^2}\\ \frac{dy}{dx}=\frac{3x^4+3x^2-6x^3-6x+6x^2+6-2x^4-2x^3+6x^2-6x+30x-30}{x^2-2x+2^2}\\ \frac{dy}{dx}=\frac{x^4+3x^2-4x^3-30x+36}{x^2-2x+2^2}\\ \frac{dy}{dx}=\frac{(x^4-4x^3+3x^2-30x+36}{x^2-2x+2^2}$

4. If $y=\frac{3x^2-5x+6}{x^2}$

Solution:

5. If $y=\frac{\sqrt{x}+10}{\sqrt{x}-10}$

Solution:

$\\\frac{dy}{dx}=\frac{d}{dx}y=\frac{d}{dx}\left [\frac{^{\sqrt{x}+10}}{^{\sqrt{x}-10}} \right ]\\ \frac{dy}{dx}=\frac{(\sqrt{x}-10)\frac{d}{dx}(\sqrt{x}+10)-(\sqrt{x}+10)\frac{d}{dx}(\sqrt{x}-10)}{(\sqrt{x}-10)^2}\\ \frac{dy}{dx}=\frac{(\sqrt{x}-10)\left (\frac{d}{dx}\sqrt{x}+\frac{d}{dx}10 \right )-(\sqrt{x}+10)\left (\frac{d}{dx}\sqrt{x}-\frac{d}{dx}10 \right )}{(\sqrt{x}-10)^2}\\ \frac{dy}{dx}=\frac{(\sqrt{x}-10)\left (\frac{1}{2\sqrt{x}}+0 \right )-(\sqrt{x}+10)\left (\frac{1}{2\sqrt{x}}-0 \right )}{(\sqrt{x}-10)^2}\\ \frac{dy}{dx}=\frac{(\sqrt{x}-10)\left (\frac{1}{2\sqrt{x}} \right )-(\sqrt{x}+10)\left (\frac{1}{2\sqrt{x}} \right )}{(\sqrt{x}-10)^2}\\$

$\\\frac{dy}{dx}=\frac{\frac{\sqrt{x}-10}{2\sqrt{x}}-\frac{\sqrt{x}+10}{2\sqrt{x}}}{(\sqrt{x}-10)^2}\\ \frac{dy}{dx}=\frac{\frac{(\sqrt{x}-10)(\sqrt{x}+10)}{2\sqrt{x}}}{(\sqrt{x}-10)^2}\\ \frac{dy}{dx}=\frac{\frac{\sqrt{x}-10-\sqrt{x}+10}{2\sqrt{x}}}{(\sqrt{x}-10)^2}\\ \frac{dy}{dx}=\frac{\frac{-20}{2\sqrt{x}}}{(\sqrt{x}-10)^2}\\ \frac{dy}{dx}=\frac{\frac{-10}{\sqrt{x}}}{(\sqrt{x}-10)^2}\\ \frac{dy}{dx}=\frac{-10}{\sqrt{x}}\div (\sqrt{x}-10)^2\\ \frac{dy}{dx}=\frac{-10}{\sqrt{x}}\times \frac{1}{(\sqrt{x}-10)^2}\\ \frac{dy}{dx}=\frac{-10}{\sqrt{x}(\sqrt{x}-10)^2}$

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Economics at Zero Marginal Cost

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Wednesday, June 10, 2020

Quotient Rule

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