Question

differentiat 1/x with respect to x from first principle​

Answer 1

The first principle of differentation states that the first derivative of a function f(x) is given by,

$\displaystyle\small\longrightarrow f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$

Here,

• $\displaystyle\smallf(x)=\dfrac{1}{x}$

Then,

$\displaystyle\small\longrightarrow f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$

$\displaystyle\small\text{ \longrightarrow f'(x)=\lim_{h\to0}\dfrac{\left(\dfrac{1}{x+h}-\dfrac{1}{x}\right)}{h} }$

$\displaystyle\small\text{ \longrightarrow f'(x)=\lim_{h\to0}\dfrac{\left(\dfrac{x-(x+h)}{x(x+h)}\right)}{h} }$

$\displaystyle\small\longrightarrow f'(x)=\lim_{h\to0}\dfrac{-h}{hx(x+h)}$

$\displaystyle\small\longrightarrow f'(x)=-\lim_{h\to0}\dfrac{1}{x(x+h)}$

$\displaystyle\small\longrightarrow f'(x)=-\dfrac{1}{x(x+0)}$

$\displaystyle\small\text{ \longrightarrow\underline{\underline{f'(x)=-\dfrac{1}{x^2}}} }$

This is the first derivative of 1/x with respect to x, derived using first principle.

Answer 2

Step-by-step explanation:

The first principle of differentation states that the first derivative of a function f(x) is given by,

\displaystyle\small\text{$\longrightarrow f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$}⟶f

′

(x)=lim

h→0

h

f(x+h)−f(x)

Here,

\displaystyle\small\text{$f(x)=\dfrac{1}{x}$}f(x)=

x

1

Then,

\displaystyle\small\text{$\longrightarrow f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$}⟶f

′

(x)=lim

h→0

h

f(x+h)−f(x)

\displaystyle\small\text{$\longrightarrow f'(x)=\lim_{h\to0}\dfrac{\left(\dfrac{1}{x+h}-\dfrac{1}{x}\right)}{h}$}⟶f

′

(x)=lim

h→0

h

(

x+h

1

−

x

1

)

\displaystyle\small\text{$\longrightarrow f'(x)=\lim_{h\to0}\dfrac{\left(\dfrac{x-(x+h)}{x(x+h)}\right)}{h}$}⟶f

′

(x)=lim

h→0

h

(

x(x+h)

x−(x+h)

)

\displaystyle\small\text{$\longrightarrow f'(x)=\lim_{h\to0}\dfrac{-h}{hx(x+h)}$}⟶f

′

(x)=lim

h→0

hx(x+h)

−h

\displaystyle\small\text{$\longrightarrow f'(x)=-\lim_{h\to0}\dfrac{1}{x(x+h)}$}⟶f

′

(x)=−lim

h→0

x(x+h)

1

\displaystyle\small\text{$\longrightarrow f'(x)=-\dfrac{1}{x(x+0)}$}⟶f

′

(x)=−

x(x+0)

1

\displaystyle\small\text{$\longrightarrow\underline{\underline{f'(x)=-\dfrac{1}{x^2}}}$}⟶

f

′

(x)=−

x

2

1

This is the first derivative of 1/x with respect to x, derived using first principle.