Economics at Zero Marginal Cost: Chain Rule - The Generalised Power Function Rule

Example

1. $y=(7x^3-5)^9\\$

Solution:

$\\\frac{dy}{dx}=\frac{d}{dx}y=\frac{d}{dx}(7x^3-5)^9\\ \frac{dy}{dx}=9(7x^3-5)^{9-1}\frac{d}{dx}(7x^3-5)\\ \frac{dy}{dx}=9(7x^3-5)^8(\frac{d}{dx}7x^3-\frac{d}{dx}5)\\ \frac{dy}{dx}=9(7x^3-5)^8(7\frac{d}{dx}x^3-0)\\ \frac{dy}{dx}=9(7x^3-5)^8(7.3x^{3-1})\\ \frac{dy}{dx}=9(7x^3-5)^8.21x^2\\ \frac{dy}{dx}=189x^2(7x^3-5)^8\\$

2. $y=(ax+b)^5$

Solution:

$\\\frac{dy}{dx}=\frac{d}{dx}y=\frac{d}{dx}(ax+b)^5\\ \frac{dy}{dx}=5(ax+b)^{5-1}\frac{d}{dx}(ax+b)\\ \frac{dy}{dx}=5(ax+b)^{4}\left (\frac{d}{dx}ax+\frac{d}{dx}b \right )\\ \frac{dy}{dx}=5(ax+b)^{4}(a\frac{d}{dx}x+0)\\ \frac{dy}{dx}=5(ax+b)^{4}(a.1)\\ \frac{dy}{dx}=5(ax+b)^{4}.a\\ \frac{dy}{dx}=5a(ax+b)^{4}\\$

3. $y=(x^2-6x)^3(3x^2+7x)^6\\$

Solution:

$\\\frac{dy}{dx}=\frac{d}{dx}y=\frac{d}{dx}(x^2-6x)^3(3x^2+7x)^6\\ \frac{dy}{dx}=(x^2-6x)^3\frac{d}{dx}(3x^2+7x)^6+(3x^2+7x)^6\frac{d}{dx}(x^2-6x)^3\\ \frac{dy}{dx}=(x^2-6x)^36(3x^2+7x)^{6-1}\frac{d}{dx}(3x^2+7x)+(3x^2+7x)^63(x^2-6x)^{3-1}\frac{d}{dx}(x^2-6x)\\ \frac{dy}{dx}=(x^2-6x)^36(3x^2+7x)^{6-1}\left ( \frac{d}{dx}3x^2+\frac{d}{dx}7x \right )+(3x^2+7x)^{6}3(x^2-6x)^{3-1}\left ( \frac{d}{dx}x^2-\frac{d}{dx}6x \right )\\ \frac{dy}{dx}=(x^2-6x)^36(3x^2+7x)^{5}\left ( 3\frac{d}{dx}x^2+7\frac{d}{dx}x \right )+(3x^2+7x)^{6}3(x^2-6x)^2(2x^{2-1}-6\frac{d}{dx}x)\\ \frac{dy}{dx}=6(x^2-6x)^3(3x^2+7x)^{5}(3.2x^{2-1}+7.1)+3(3x^2+7x)^6(x^2-6x)^2(2x-6.1)\\ \frac{dy}{dx}=6(x^2-6x)^3(3x^2+7x)^{5}(6x+7)+3(3x^2+7x)^6(x^2-6x)^2(2x-6)\\$

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Economics at Zero Marginal Cost

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Wednesday, June 10, 2020

Chain Rule - The Generalised Power Function Rule

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