Question

Find the derivative by the limit process.
f(x)=1-x^2

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:f(x) = 1 - {x}^{2}$

So,

$\rm :\longmapsto\:f(x + h) = 1 - {(x + h)}^{2}$

Using definition, we have

$\rm :\longmapsto\:f'(x) = \lim_{h \to 0}\dfrac{f(x + h) - f(x)}{h}$

$\rm \: \: = \: \lim_{h \to 0}\dfrac{\bigg(1 - {(x + h)}^{2} \bigg) - (1 - {x}^{2})}{h}$

$\rm \: \: = \: \lim_{h \to 0}\dfrac{1 - {(x + h)}^{2} - 1 + {x}^{2} }{h}$

$\rm \: \: = \: \lim_{h \to 0}\dfrac{ - {(x + h)}^{2} + {x}^{2} }{h}$

$\rm \: \: = \: \lim_{h \to 0}\dfrac{ - {(x }^{2} + {h}^{2} + 2hx) + {x}^{2} }{h}$

$\rm \: \: = \: \lim_{h \to 0}\dfrac{ - {x }^{2} - {h}^{2} - 2hx+ {x}^{2} }{h}$

$\rm \: \: = \: \lim_{h \to 0}\dfrac{ - {h}^{2} - 2hx}{h}$

$\rm \: \: = \: \lim_{h \to 0}\dfrac{-h( {h} + 2x)}{h}$

$\rm \: \: = - \: \lim_{h \to 0}(h + 2x)$

$\rm \: \: = \: - (0 + 2x)$

$\rm \: \: = \: - 2x$

$\bf\implies \:\dfrac{d}{dx}(1 - {x}^{2}) = - \: 2x$

Additional Information :-

$\green{\boxed{ \bf{ \:\dfrac{d}{dx}k = 0}}}$

$\green{\boxed{ \bf{ \:\dfrac{d}{dx}x = 1}}}$

$\green{\boxed{ \bf{ \:\dfrac{d}{dx} {e}^{x} = {e}^{x} }}}$

$\green{\boxed{ \bf{ \:\dfrac{d}{dx} {a}^{x} = {a}^{x} log(a)}}}$

$\green{\boxed{ \bf{ \:\dfrac{d}{dx}logx = \dfrac{1}{x}}}}$

$\green{\boxed{ \bf{ \:\dfrac{d}{dx} \sqrt{x} = \dfrac{1}{2 \sqrt{x} }}}}$

$\green{\boxed{ \bf{ \:\dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}}}$