Economics at Zero Marginal Cost: Chain Rule - The Generalised Logarithmic Rule

Excercise:

Find $\frac{dy}{dx}$

1. $y=log\;7x^2$

Solution:

$\\\frac{dy}{dx}=\frac{d}{dx}y=\frac{d}{dx}log\;7x^2\\ \frac{dy}{dx}=\frac{1}{7x^2}\frac{d}{dx}7x^2\\ \frac{dy}{dx}=\frac{1}{7x^2}.7\frac{d}{dx}x^2\\ \frac{dy}{dx}=\frac{1}{7x^2}.7.2x^{2-1}\\ \frac{dy}{dx}=\frac{1}{7x^2}.14x\\ \frac{dy}{dx}=\frac{1}{x^2}2x\\ \frac{dy}{dx}=\frac{2}{x}$

2. $y=log\left (\frac{x^2+5}{x^2-5} \right )$

Solution:

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

$\\\frac{dy}{dx}=\frac{\frac{1}{x^2+5}}{x^2-5}\left ( \frac{(x^2-5).2x-(x^2+5).2x}{(x^2-5)^2} \right )\\ \frac{dy}{dx}=\frac{\frac{1}{x^2+5}}{x^2-5}\left ( \frac{(2x^3-10x)-(2x^3+10x)}{(x^2-5)^2} \right )\\ \frac{dy}{dx}=\frac{\frac{1}{x^2+5}}{x^2-5}\left ( \frac{2x^3-10x-2x^3+10x}{(x^2-5)^2} \right )\\ \frac{dy}{dx}=\frac{\frac{1}{x^2+5}}{x^2-5}\left ( \frac{-20x}{(x^2-5)^2} \right )\\ \frac{dy}{dx}=\frac{\frac{-20x}{(x^2+5)^2}}{\frac{x^2+5}{x^2-5}}\\ \frac{dy}{dx}=\frac{-20x}{(x^2+5)^2}\div \frac{(x^2+5)}{(x^2-5)}\\ \frac{dy}{dx}=\frac{-20x}{(x^2+5)^2}\times \frac{(x^2-5)}{(x^2+5)}\\ \frac{dy}{dx}=\frac{-20x}{(x^2-5)(x^2+5)}\\ \frac{dy}{dx}=\frac{-20x}{(x^2)^2-5^2}\\ \frac{dy}{dx}=\frac{-20x}{x^4-25}$

3. $y=log \left [\frac{4x}{2+x} \right ]$

Solution:

$\\\frac{dy}{dx}=\frac{d}{dx}y=\frac{d}{dx}log\left [\frac{4x}{2+x} \right ]\\ \frac{dy}{dx}=\frac{\frac{1}{4x}}{2+x}\frac{d}{dx}\left (\frac{4x}{2+x} \right )\\ \frac{dy}{dx}=\frac{\frac{1}{4x}}{2+x}\left ( \frac{(2+x)\frac{d}{dx}4x-4x\frac{d}{dx}(2+x)}{(2+x)^2} \right )\\ \frac{dy}{dx}=\frac{\frac{1}{4x}}{2+x}\left ( \frac{(2+x)4\frac{d}{dx}x-4x\left ( \frac{d}{dx}2+\frac{d}{dx}x \right )}{2+x} \right )\\ \frac{dy}{dx}=\frac{\frac{1}{4x}}{2+x}\left ( \frac{(2+x)4.1-4x(0+1)}{(2+x)} \right )\\ \frac{dy}{dx}=\frac{\frac{1}{4x}}{2+x}\left ( \frac{4(2+x)-4x(1)}{(2+x)^2} \right )\\ \frac{dy}{dx}=\frac{\frac{1}{4x}}{2+x}\left ( \frac{(8+4x)-4x}{(2+x)^2} \right )\\ \frac{dy}{dx}=\frac{\frac{1}{4x}}{2+x}\left ( \frac{8+4x-4x}{(2+x)^2} \right )\\ \frac{dy}{dx}=\frac{\frac{1}{4x}}{2+x}\left ( \frac{8}{(2+x)} \right )\\ \frac{dy}{dx}=\frac{\frac{8}{(2+x)^2}}{\frac{4x}{2+x}}\\ \frac{dy}{dx}=\frac{8}{(2+x)^2}\div \frac{4x}{2+x}\\ \frac{dy}{dx}=\frac{8}{(2+x)^2}\times \frac{2+x}{4x}\\ \frac{dy}{dx}=\frac{2}{x(2+x)}\\$

4. $f(x)= log(x-\sqrt{6-x}+2)$

Solution:

$\\\frac{dy}{dx}=\frac{d}{dx}y=\frac{d}{dx}log(x-\sqrt{6-x}+2)\\ \frac{dy}{dx}=\frac{1}{x-\sqrt{6-x}+2}\frac{d}{dx}(x-\sqrt{6-x}+2)\\ \frac{dy}{dx}=\frac{1}{x-\sqrt{6-x}+2}\left ( \frac{d}{dx}x-\frac{d}{dx}\sqrt{6-x}+\frac{d}{dx}2 \right )\\ \frac{dy}{dx}=\frac{1}{x-\sqrt{6-x}+2}(1-\frac{1}{2\sqrt{6-x}}\frac{d}{dx}(6-x)+0)\\ \frac{dy}{dx}=\frac{1}{x-\sqrt{6-x}+2}(1-\frac{1}{2\sqrt{6-x}}\left ( \frac{d}{dx}6-\frac{d}{dx}x \right ))\\ \frac{dy}{dx}=\frac{1}{x-\sqrt{6-x}+2}(1-\frac{1}{2\sqrt{6-x}}(0-1)\\ \frac{dy}{dx}=\frac{1}{x-\sqrt{6-x}+2}(1-\frac{1}{2\sqrt{6-x}}(-1))\\ \frac{dy}{dx}=\frac{1}{x-\sqrt{6-x}+2}(1+\frac{1}{2\sqrt{6-x}})\\ \frac{dy}{dx}=\frac{1+\frac{1}{2\sqrt{6-x}}}{x-\sqrt{6-x}+2}$

5. $y=logx+\sqrt{x^2+a^2}$

Solution:

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

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Economics at Zero Marginal Cost

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Wednesday, June 10, 2020

Chain Rule - The Generalised Logarithmic Rule

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