Economics at Zero Marginal Cost: Chain Rule - Differentiation of roots of Expression

Excercise:

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

1. $y=\sqrt{4x+x^2}$

Solution:

$\\\frac{dy}{dx}=\frac{d}{dx}y=\frac{d}{dx}\sqrt{4x+x^2}\\ \frac{dy}{dx}=\frac{1}{2\sqrt{4x+x^2}}.\frac{d}{dx}(4x+x^2)\\ \frac{dy}{dx}=\frac{1}{2\sqrt{4x+x^2}}.\left ( \frac{d}{dx}4x+\frac{d}{dx}x^2 \right )\\ \frac{dy}{dx}=\frac{1}{2\sqrt{4x+x^2}}.\left ( 4\frac{d}{dx}x+2x^{2-1} \right )\\ \frac{dy}{dx}=\frac{1}{2\sqrt{4x+x^2}}.(4.1+2x)\\ \frac{dy}{dx}=\frac{1}{2\sqrt{4x+x^2}}.(4+2x)\\ \frac{dy}{dx}=\frac{(4+2x)}{2\sqrt{4x+x^2}}\\ \frac{dy}{dx}=\frac{x+2}{\sqrt{4x+x^2}}\$

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

Solution:

$\;\\y=\sqrt{\frac{5-x}{5+x}}\\ \frac{dy}{dx}=\frac{d}{dx}y=\frac{d}{dx}\frac{\sqrt{5-x}}{\sqrt{5+x}}\\ \frac{dy}{dx}=\frac{\sqrt{5+x}\frac{d}{dx}\sqrt{5-x}-\sqrt{5-x}\frac{d}{dx}\sqrt{5+x}}{(\sqrt{5+x})^2}\\ \frac{dy}{dx}=\frac{(\sqrt{5+x})\frac{1}{2\sqrt{5-x}\frac{d}{dx}}(5-x)-(\sqrt{5-x})\frac{1}{2\sqrt{5+x}}\frac{d}{dx}(5+x)}{5+x}\\ \frac{dy}{dx}=\frac{\frac{\sqrt{5+x}}{2\sqrt{5-x}}\left ( \frac{d}{dx}5-\frac{d}{dx}x \right )-\frac{\sqrt{5-x}}{2\sqrt{5+x}}(\frac{d}{dx}5+\frac{d}{dx}x)}{5+x}\\ \frac{dy}{dx}=\frac{\frac{\sqrt{5+x}}{2\sqrt{5-x}}(0-1)-\frac{\sqrt{5-x}}{2\sqrt{5+x}}(0+1)}{5+x}\\ \frac{dy}{dx}=\frac{\frac{\sqrt{5+x}}{2\sqrt{5-x}}(-1)-\frac{\sqrt{5-x}}{2\sqrt{5+x}}(1)}{5+x}\\ \frac{dy}{dx}=\frac{\frac{\sqrt{5+x}}{2\sqrt{5-x}}-\frac{\sqrt{5-x}}{2\sqrt{5+x}}}{5+x}\\$

3. $y=\frac{x^2}{\sqrt{1+x^2}}$

Solution:

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

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

Solution:

$\\\frac{dy}{dx}=\frac{d}{dx}y=\frac{d}{dx}\sqrt{x^3-5x+3}\\ \frac{dy}{dx}=\frac{1}{2\sqrt{x^3-5x+3}}\frac{d}{dx}(x^3-5x+3)\\ \frac{dy}{dx}=\frac{1}{2\sqrt{x^3-5x+3}}\left ( \frac{d}{dx}x^3-\frac{d}{dx}5x+\frac{d}{dx}3 \right )\\ \frac{dy}{dx}=\frac{1}{2\sqrt{x^3-5x+3}}\left( 3x^{3-1}-5\frac{d}{dx}+0 \right)\\ \frac{dy}{dx}=\frac{1}{2\sqrt{x^3-5x+3}}(3x^2-5.1)\\ \frac{dy}{dx}=\frac{1}{2\sqrt{x^3-5x+3}}(3x^2-5)\\ \frac{dy}{dx}=\frac{(3x^2-5)}{2\sqrt{x^3-5x+3}}$

Economics at Zero Marginal Cost

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Wednesday, June 10, 2020

Chain Rule - Differentiation of roots of Expression

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