Economics at Zero Marginal Cost: Product Rule

The derivative of the product of two differentiable functions is equal to the first function times the derivative of the second plus the second function times the derivative of the second function:

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

Excercise:

Find $\frac{dy}{dx}$

$\\\1.\;y=(1-\sqrt{x})(1+\sqrt{x})\\ \2.\;y=(x^2+4x)(x^2-4x)\\ \3.\;y=(ax^2+bx+cx)(px+q)\\ \4.\;y=(4x^2+2x)(8x^3-3x^2)$

Excercise:

Find $\frac{dy}{dx}$

1. $y=(1-\sqrt{x})(1+\sqrt{x})\\$

Solution:

2. If $y=(x^2+4x)(x^2-4x)$

Solution:

$\\\frac{dy}{dx}=\frac{d}{dx}y=\frac{d}{dx}[(x^2+4x)(x^2-4x)]\\ \frac{dy}{dx}=(x^2+4x)\frac{d}{dx}(x^2-4x)+(x^2-4x)\frac{d}{dx}(x^2+4x)\\ \frac{dy}{dx}=(x^2+4x)\left ( \frac{d}{dx}x^2-\frac{d}{dx}4x \right )+(x^2-4x)\left ( \frac{d}{dx}x^2+\frac{d}{dx}4x \right )\\ \frac{dy}{dx}=(x^2+4x)\left (2x^{2-1}-4\frac{d}{dx}x \right )+(x^2-4x)\left (2x^{2-1}+4\frac{d}{dx}x \right )\\ \frac{dy}{dx}=(x^2+4x^2)(2x-4.1)+(x^2-4x)(2x+4.1)\\ \frac{dy}{dx}=(x^2+4x^2)(2x-4)+(x^2-4x)(2x+4)\\ \frac{dy}{dx}=(2x^3-4x^2+8x^2-16x)+(2x^3+4x^2-8x^2-16x)\\ \frac{dy}{dx}=2x^3-4x^2+8x^2-16x+2x^3+4x^2-8x^2-16x\\ \frac{dy}{dx}=4x^3-32x\\ \frac{dy}{dx}=4x(x^2-8)$

3.If $y=(ax^2+bx+cx)(px+q)\\$

Solution:

$\\\frac{dy}{dx}=\frac{d}{dx}y=\frac{d}{dx}[(ax^2+bx+cx)(px+q)]\\ \frac{dy}{dx}=(ax^2+bx+cx)\frac{d}{dx}(px+q)+(px+q)\frac{d}{dx}(ax^2+bx+cx)\\ \frac{dy}{dx}=(ax^2+bx+cx)\left ( \frac{d}{dx}px+\frac{d}{dx}q \right )+(px+q)\left ( \frac{d}{dx}ax^2+\frac{d}{dx}bx+\frac{d}{dx}cx \right )\\ \frac{dy}{dx}=(ax^2+bx+cx)\left (p\frac{d}{dx}x+0 \right )+(px+q)\left ( a\frac{d}{dx}x^2+b\frac{d}{dx}x+c\frac{d}{dx}x \right ) \\ \frac{dy}{dx}=(ax^2+bx+cx)(p.1)+(px+q)(a2x^{2-1}+b.1+c.1)\\ \frac{dy}{dx}=(ax^2+bx+cx)p+(px+q)(2ax+b+c)\\$

4. If $y=(4x^2+2x)(8x^3-3x^2)$

Solution:

$\\\frac{dy}{dx}=\frac{d}{dx}[(4x^2+2x)(8x^3-3x^2)]\\ \frac{dy}{dx}=(4x^2+2x)\frac{d}{dx}(8x^3-3x^2)+(8x^3-3x^2)\frac{d}{dx}(4x^2+2x)\\ \frac{dy}{dx}=(4x^2+2x)\left (\frac{d}{dx}8x^3-\frac{d}{dx}3x^2\right )+(8x^3-3x^2)\left (\frac{d}{dx}4x^2-\frac{d}{dx}2x\right )\\ \frac{dy}{dx}=(4x^2+2x)\left (8\frac{d}{dx}x^3-3\frac{d}{dx}x^2\right )+(8x^3-3x^2)\left (4\frac{d}{dx}x^2+2\frac{d}{dx}x\right )\\ \frac{dy}{dx}=(4x^2+2x)(8.3x^{3-1}-3.2x^{2-1})+(8x^3-3x^2)(4.2x^{2-1}+2.1)\\ \frac{dy}{dx}=(4x^2+2x)(24x^2-6x)+(8x^3-3x^2)(8x+2)\\ \frac{dy}{dx}=(96x^4-24x^3+48x^3-12x^2)+(64x^4+16x^3-24x^3-6x^2)\\ \frac{dy}{dx}=96x^4-24x^3+48x^3-12x^2+64x^4+16x^3-24x^3-6x^2\\ \frac{dy}{dx}=160x^4-48x^3+64x^3-18x^2\\ \frac{dy}{dx}=160x^4+16x^3-18x^2\\ \frac{dy}{dx}=2x^2(80x^2+8x-9)$

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Economics at Zero Marginal Cost

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Wednesday, June 10, 2020

Product Rule

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