Question

Find the derivative of the following from the first principle
1. x²-27
2. (x-1)(x-2)​

Answer 1

$\huge\sf\pink{\underline{\underline{Solution}}}\::$

$\rule{170}2$

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$\sf\gray{(i) \ Let \ f(x) = x^3 - 27. \ According ,\ from \ the \ first \ principle}$

$\sf\red{f'(x) \:=\:lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}}$

⠀⠀⠀ $\sf\orange{=\:lim_{h\to 0}\dfrac{{\large [} (x+h)^3 -27{\large ]} - (x^3 - 27)}{h}}$

⠀⠀⠀ $\sf\green{=\:lim_{h\to 0}\dfrac{x^3 + h^3 + 3x^2 h + 3xh^2 - x^3}{h}}$

⠀⠀⠀ $\sf\blue{=\:lim_{h\to 0}\dfrac{h^3 + 3x^2 h + 3xh^2}{h}}$

⠀⠀⠀ $\sf\purple{=\:lim_{h\to 0} (h^2 + 3x^2 + 3xh)}$

⠀⠀⠀ $\sf\red{=\:0+ 3x^2 + 0}$

⠀⠀⠀ $\sf\orange{=\:3x^2}$

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$\rule{170}2$

⠀

$\sf\gray{(i) \ Let \ f(x) = (x-1)(x-2). \ According ,\ from \ the \ first \ principle}$

$\sf\red{f'(x) \:=\:lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}}$

⠀⠀⠀ $\sf\orange{=\:lim_{h\to 0}\dfrac{(x+h-1)(x+h-2)-(x-1)(x-2)}{h}}$

⠀⠀⠀ $\sf\green{=\:lim_{h\to 0}\dfrac{(x^2 +hx -2x + hx + h^2 - 2h - h + 2)-(x^2 - 2x - x+2)}{h}}$

⠀⠀⠀ $\sf\blue{=\:lim_{h\to 0}\dfrac{(hx+hx+h^2 -2h-h)}{h}}$

⠀⠀⠀ $\sf\purple{=\:lim_{h\to 0}\dfrac{2hx+h^2 - 3h}{h}}$

⠀⠀⠀ $\sf\red{=\:lim_{h\to 0}(2x+h-3)}$

⠀⠀⠀ $\sf\orange{=\:(2x+0-3)}$

⠀⠀⠀ $\sf\green{=\:2x-3}$

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$\rule{170}2$

Answer 2

1. 2x

2. 2x-3....... ..