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

Excercise:

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

1. $y=5e^{2-x^2}$

Solution:

$\\\frac{dy}{dx}=\frac{d}{dx}y=\frac{d}{dx}5e^{2-x^2}\\ \frac{dy}{dx}=5\frac{d}{dx}e^{2-x^2}\\ \frac{dy}{dx}=5e^{2-x^2}\frac{d}{dx}(2-x^2)\\ \frac{dy}{dx}=5e^{2-x^2}\left ( \frac{d}{dx}2-\frac{d}{dx}x^2 \right )\\ \frac{dy}{dx}=5e^{2-x^2}(0-2x^{2-1})\\ \frac{dy}{dx}=5e^{2-x^2}.-2x\\ \frac{dy}{dx}=-10xe^{2-x^2}.\\$

2. $y=e^{ax3+bx^2+cx+d}$

Solution:

$\\\frac{dy}{dx}=\frac{d}{dx}y=\frac{d}{dx}e^{ax3+bx^2+cx+d}\\ \frac{dy}{dx}=e^{ax3+bx^2+cx+d}\frac{d}{dx}(ax3+bx^2+cx+d)\\ \frac{dy}{dx}=e^{ax3+bx^2+cx+d}(\frac{d}{dx}ax^3+\frac{d}{dx}bx^2+\frac{d}{dx}cx+\frac{d}{dx}d)\\ \frac{dy}{dx}=e^{ax3+bx^2+cx+d}(a\frac{d}{dx}x^3+b\frac{d}{dx}x^2+c\frac{d}{dx}x+0)\\ \frac{dy}{dx}=e^{ax3+bx^2+cx+d}(a3x^{3-1}+b3x^{2-1}+c.1)\\ \frac{dy}{dx}=e^{ax3+bx^2+cx+d}(3ax^2+2bx+c)\\ \frac{dy}{dx}=(3ax^2+2bx+c)e^{ax3+bx^2+cx+d}\\$

3. $y=e\sqrt{(1+x^3)}$

Solution:

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

4. $f(x)=e^{x3-^{\sqrt{3-1}}+1}$

Solution:

Let $y=f(x)=e^{x3-^{\sqrt{3-1}+1}}$

$\\\frac{dy}{dx}=\frac{d}{dx}y=\frac{d}{dx}e^{x3-^{\sqrt{3-x}+1}}\\ \frac{dy}{dx}=e^{x3-^{\sqrt{3-x}+1}}\frac{d}{dx}(x3-\sqrt{3-x}+1)\\ \frac{dy}{dx}=e^{x3-^{\sqrt{3-x}+1}}\left ( \frac{d}{dx}x^3-\frac{d}{dx}\sqrt{3-x}+\frac{d}{dx}1 \right )\\ \frac{dy}{dx}=e^{x3-^{\sqrt{3-x}+1}}(3x^{3-1}-\frac{1}{2\sqrt{3-x}}\frac{d}{dx}(3-x)+0)\\ \frac{dy}{dx}=e^{x3-^{\sqrt{3-x}+1}}(3x^2-\frac{1}{2\sqrt{3-x}}\left ( \frac{d}{dx}3-\frac{d}{dx}x \right ))\\ \frac{dy}{dx}=e^{x3-^{\sqrt{3-x}+1}}(3x^2-\frac{1}{2\sqrt{3-x}}(0-1))\\ \frac{dy}{dx}=e^{x3-^{\sqrt{3-x}+1}}(3x^2-\frac{1}{2\sqrt{3-x}}(-1)\\ \frac{dy}{dx}=e^{x3-^{\sqrt{3-x}+1}}(3x^2+\frac{1}{2\sqrt{3-x}})\\ \frac{dy}{dx}=\left ( 3x^2+\frac{1}{2\sqrt{3-x}} \right )e^{x3-^{\sqrt{3-x}+1}}$

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Economics at Zero Marginal Cost

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Wednesday, June 10, 2020

Chain Rule - The Generalised Exponential Rule:

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