Question

Find using first principle the derivative of x-1÷2x+7

Answer 1

Answer:

9/(2x+7)^2

Step-by-step explanation:

use u/v formula

Answer 2

The derivative of f(x)

$f'(x)=\frac{9}{(2x+7)^2}$

Step-by-step explanation:

$f(x)=\frac{x-1}{2x+7}$

$f(x+h)=\frac{x+h-1}{2(x+h)+7}$

By first principle

$f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{x+h-1}{2x+2h+7}-\frac{x-1}{2x+7}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{(x+h-1)(2x+7)-(x-1)(2x+2h+7)}{h(2x+2h+7)(2x+7)}$

$f'(x)=lim_{h\rightarrow 0}\frac{2x^2+7x+2xh+7h-2x-7-2x^2-2xh-7x+2x+2h+7}{h(2x+7)(2x+2h+7)}$

$f'(x)=lim_{h\rightarrow 0}\frac{9h}{h(2x+7)(2x+2h+7)}=\frac{9}{(2x+7)^2}$

Hence, the derivative of f(x)

$f'(x)=\frac{9}{(2x+7)^2}$

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