Question

Differentiate 2x+3/3x+2 by using first principles

Answer 1

is it right. ??.....I hope this will help you

Answer 2

Answer:

The differentiation of given function by first principle is

$f'(x) = \frac{ - 5}{9 {x}^{2} + 12x + 4 }$

Step-by-step explanation:

Let the given function be

$f(x) = \frac{2x + 3}{3x + 2}$

According to first principle of differentiation,the differential of a function f(x) is f'(x) and it is calculated as shown below

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

As we know f(x),then f(x+h) would be

$f(x + h) = \frac{2(x + h) + 3}{3(x + h) + 2} = \frac{2x + 3 + 2h}{3x + 2 + 3h}$

Substituting this in the first principle formula,we get

$f'(x)=Lim_{h->0} \frac{ \frac{2x + 3 + 2h}{3x + 2 + 3h} - \frac{2x + 3}{3x + 2} }{h} \\ = Lim_{h->0} \frac{ \frac{(2x + 3 + 2h)(3x + 2) - (2x + 3)(3x + 2 + 3h)}{(3x + 2)(3x + 2 + 3h)} }{h} \\ = Lim_{h->0} \frac{ \frac{(6x {}^{2} + 13x + 6 + 6hx + 4h) - (6x {}^{2} + 13x + 6xh + 6 + 9h)}{9x {}^{2} + 4 + 12x + 9xh + 6h } }{h} \\ = Lim_{h->0} \frac{ \frac{ - 5h}{9 {x}^{2} + 12x + 9xh + 6h + 4} }{h} \\ = Lim_{h->0} \frac{ - 5}{9 {x}^{2} + 12x + 9xh + 6h + 4 }$

substituting the value of h in the limit,we get

$= > Lim_{h->0} \frac{ - 5}{9 {x}^{2} + 12x + 9x(0) + 6(0) + 4 } \\ = \frac{ - 5}{9 {x}^{2} + 12x + 4}$

Therefore the differentiation of the function by first principle is

$f'(x) = \frac{ - 5}{9 {x}^{2} + 12x + 4 }$

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