Question

Find the derivative by first principle:
$\rm \frac{x-1}{2x+7}$

Answer 1

Answer:

$\mathbf{\frac{d}{dx}[\frac{x\ -\ 1}{2x\ +\ 7}]\ =\ \frac{9}{(2x\ +\ 7)^{2}}}$

Step-by-step explanation:

In this question,

We need to find the derivative of $\frac{x\ -\ 1}{2x\ +\ 7}$

According to the question,

$\frac{\mathrm{d} }{\mathrm{d} x}[\frac{x\ -\ 1}{2x\ +\ 7}]$

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

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

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

=>  f'(x)  = $Lim_{h-0}\frac{((2x²+2hx -2x + 7x + 7h - 7)  - (2x² -2x +2hx - 2h + 7x - 7))}{(h(2x + 2h + 7)(2x + 7))}$

=>  f'(x)  = $Lim_{h-0} \frac{(9h)}{(h(2x + 2h + 7)(2x + 7))}$

=>  f'(x)  = $Lim_{h-0} \frac{(9)}{((2x + 2h + 7)(2x + 7))}</p><p></p><p>putting h = 0</p><p></p><p>=> f'(x)  = [tex]\frac{9}{(2x + 7)^{2}}$

On solving,

$\frac{d}{dx}[\frac{x\ -\ 1}{2x\ +\ 7}]\ =\ \frac{9}{(2x\ +\ 7)^{2}}$

Hence, $\mathbf{\frac{d}{dx}[\frac{x\ -\ 1}{2x\ +\ 7}]\ =\ \frac{9}{(2x\ +\ 7)^{2}}}$

Answer 2

Answer:

9/(2x + 7)²

Step-by-step explanation:

f(x)  = (x - 1) / (2x + 7)

f'(x)  = Lim h- 0  (f(x+h) - f(x) ) / h

=> f'(x)  = Lim h-0   ((x + h - 1) /(2x + 2h + 7)   -  (x - 1) / (2x + 7))/h

=> f'(x)  = Lim h-0  ( (x + h - 1) (2x + 7)  - (x - 1)(2x + 2h + 7))/(h(2x + 2h + 7)(2x + 7))

=>  f'(x)  = Lim h-0  ((2x²+2hx -2x + 7x + 7h - 7)  - (2x² -2x +2hx - 2h + 7x - 7))/(h(2x + 2h + 7)(2x + 7))

=>  f'(x)  = Lim h-0  (9h)/(h(2x + 2h + 7)(2x + 7))

=>  f'(x)  = Lim h-0  (9)/((2x + 2h + 7)(2x + 7))

putting h = 0

=> f'(x)  = 9/(2x + 7)²