Economics at Zero Marginal Cost: Sum & Difference Rule

Sum and Difference Rule:

The derivative of a sum (difference) of two functions is the sum (difference) of the derivatives of the two functions:

$\frac{d}{dx}[f(x)\pm g(x)]=\frac{d}{dx}f(x)\pm\frac{d}{dx}g(x)=f'(x)\pm\;g'(x)$

Excercise:

Find $\frac{dy}{dx}$

$\\\1.\;y=3x^4+x^2-1\\ \2.\;y=3x^3+7x^2-9x+20\\ \3.\;y=ax^2+bx+c\\ \4.\;y=3x^2-4x^{\frac{1}{3}}+x^{-\frac{1}{2}}\\ \5.\;y=5x^4+3x^3-6\sqrt{x}+10\\ \6.\;y=7x^{1/2}+\frac{3}{x}-\frac{5}{3\sqrt{x}}$

Excercise:

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

1. If $y=3x^4+x^2-1$

Solution:

$\\\frac{dy}{dx}=\frac{d}{dx}y=\frac{d}{dx}(3x^4+x^2-1)\\ \frac{dy}{dx}=\frac{d}{dx}3x^4+\frac{d}{dx}x^2-\frac{d}{dx}1\\ \frac{dy}{dx}=3\frac{d}{dx}x^4+2x^{2-1}-0\\ \frac{dy}{dx}=3.4x^{4-1}+2x\\ \frac{dy}{dx}=12x^3+2x$

2. If $y=3x^3+7x^2-9x+20$

Solution:

$\\\frac{dy}{dx}=\frac{d}{dx}y=\frac{d}{dx}(3x^3+7x^2-9x+20)\\ \frac{dy}{dx}=\frac{d}{dx}3x^3+\frac{d}{dx}7x^2-\frac{d}{dx}9x+\frac{d}{dx}20\\ \frac{dy}{dx}=3\frac{d}{dx}x^3+7\frac{d}{dx}x^2-9\frac{d}{dx}x+0\\ \frac{dy}{dx}=3.3x^{3-1}+7.2x^{2-1}-9.1\\ \frac{dy}{dx}=9x^2+14x-9\\$

3. If $y=ax^2+bx+c\\$

Solution:

$\\\frac{dy}{dx}=\frac{d}{dx}y=\frac{d}{dx}(ax^2+bx+c)\\ \frac{dy}{dx}=\frac{d}{dx}ax^2+\frac{d}{dx}bx+\frac{d}{dx}c\\ \frac{dy}{dx}=a\frac{d}{dx}x^2+b\frac{d}{dx}x+0\\ \frac{dy}{dx}=a.2x^{2-1}+b.1\\ \frac{dy}{dx}=2ax+b\\$

4. If $y=3x^2-4x^{\frac{1}{3}}+x^{-\frac{1}{2}}$

Solution:

$\\\frac{dy}{dx}=\frac{d}{dx}y=\frac{d}{dx}(3x^2-4x^{\frac{1}{3}}+x^{-\frac{1}{2}})\\ \frac{dy}{dx}=\frac{d}{dx}3x^2-\frac{d}{dx}4x^{\frac{1}{3}}+\frac{d}{dx}x^{-\frac{1}{2}}\\ \frac{dy}{dx}=3\frac{d}{dx}x^2-4\frac{d}{dx}x^{\frac{1}{3}}+\left ( -\frac{1}{2} \right )x^{-\frac{1}{2}-1}\\ \frac{dy}{dx}=3.2x^{2-1}-4.\frac{1}{3}x^{\frac{1}{3}-1}-\frac{1}{2}x^{-\frac{3}{2}}\\ \frac{dy}{dx}=6x-\frac{4}{3}x^{-\frac{2}{3}}-\frac{1}{2}x^{-\frac{3}{2}}\\ \frac{dy}{dx}=6x-\frac{4}{3x^{2/3}}-\frac{1}{2x^{3/2}}$

5. If $y=5x^4+3x^3-6\sqrt{x}+10$

Solution:

$\\\frac{dy}{dx}=\frac{d}{dx}y=\frac{d}{dx}(5x^4+3x^3-6\sqrt{x}+10)\\ \frac{dy}{dx}=\frac{d}{dx}5x^4+\frac{d}{dx}3x^3-\frac{d}{dx}6\sqrt{x}+\frac{d}{dx}10\\ \frac{dy}{dx}=5\frac{d}{dx}x^4+3\frac{d}{dx}x^3-6\frac{d}{dx}\sqrt{x}+0\\ \frac{dy}{dx}=5.4x^{4-1}+3.3x^{3-1}-6.\frac{1}{2\sqrt{x}}\\ \frac{dy}{dx}=20x^3+9x^2-\frac{3}{\sqrt{x}}\\$

6. If $y=7x^{1/2}+\frac{3}{x}-\frac{5}{3\sqrt{x}}$

Solution:

$\\\frac{dy}{dx}=\frac{d}{dx}y=\frac{d}{dx}(7x^{\frac{1}{2}}+\frac{3}{x}-\frac{5}{3\sqrt{x}})\\ \frac{dy}{dx}=\frac{d}{dx}7x^{\frac{1}{2}}+\frac{d}{dx}\frac{3}{x}-\frac{d}{dx}\frac{5}{3\sqrt{x}}\\ \frac{dy}{dx}=7\frac{d}{dx}x^{\frac{1}{2}}+\frac{d}{dx}3x^{-1}-\frac{d}{dx}\frac{5}{3x^{1/2}}\\ \frac{dy}{dx}=7.\frac{1}{2}x^{\frac{1}{2}-1}+3\frac{d}{dx}x^{-1}-\frac{d}{dx}\frac{5}{3}x^{-\frac{1}{2}}\\ \frac{dy}{dx}=\frac{7}{2}x^{-\frac{1}{2}}+3.-1x^{-1-1}-\frac{5}{3}.\frac{d}{dx}x^{-\frac{1}{2}}\\ \frac{dy}{dx}=\frac{7}{2}x^{-\frac{1}{2}}+3.-1x^{-1-1}-\frac{5}{3}.\frac{d}{dx}x^{-\frac{1}{2}}\\ \frac{dy}{dx}=\frac{7}{2x^{1/2}}-3x^{-2}-\frac{5}{3}.-\frac{1}{2}x^{-\frac{1}{2}-1}\\ \frac{dy}{dx}=\frac{7}{2x^{1/2}}-\frac{3}{x^2}+\frac{5}{6}x^{-\frac{3}{2}}\\ \frac{dy}{dx}=\frac{7}{2x^{1/2}}-\frac{3}{x^2}+\frac{5}{6x^{3/2}}$

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Economics at Zero Marginal Cost

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Wednesday, June 10, 2020

Sum & Difference Rule

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