Question

Find the derivatives of
√x and ㏒x using first principle

Answer 1

Answer:

Step-by-step explanation:

Let y = f(x) be a function . If h or (Δx) be a small increment in x and the corresponding increment in y or f(x) be Δy = f( x + h ) then the derivative of f(x) is defined as

• $\boxed{\lim_{h\to 0} = \frac{f(x+h) - f(x)}{h}}$

Let f(x) = $\sqrt{x}$

then f( x + h ) = $\sqrt{x + h}$

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

$\lim_{h\to 0} = \frac{\sqrt{x + h} - \sqrt{x} }{h}$

$\lim_{h\to 0} = \frac{\sqrt{(x + h}-\sqrt{x} )\sqrt{(x + h})+\sqrt{x} }{h(\sqrt{x+h}+\sqrt{x} }$

$\lim_{h\to 0} = \frac{x + h - x}{h(\sqrt{x + h}+\sqrt{x} }$

$\lim_{h\to 0} = \frac{1}{\sqrt{x + h}+\sqrt{x} }$

$\boxed{ \frac{1}{2\sqrt{x} }}$

Thus derivative of $\sqrt{x}$ = $\frac{1}{2\sqrt{x} }$

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• log (x)

f(x) = log x

f(x + h) = log (x + h)

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

$\lim_{h\to 0} = \frac{log(x + h)-log x}{h}$

$\lim_{h\to 0} = \frac{log(\frac{x + h}{x}) }{h}$

$\lim_{h\to 0} = \frac{1}{h} [log(1+\frac{h}{x} )]$

Let $\frac{h}{x}$ = t

i.e. h = tx

$f'(x) = \lim_{h\to 0} \frac{1}{tx}log (1 + t) = \frac{1}{x}$

$\boxed{ \frac{1}{x}}$

Thus derivative of log (x) = $= \frac{1}{x}$

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