Question

a) Find the derivative of x ^ 2 using the first principle of derivative.​

Answer 1

Answer:

x²=√X

HOPE ITS HELPFULL

Answer 2

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

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

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

Now,

Using definition of first principle of differentiation,

$\rm :\longmapsto\:f'(x) = \displaystyle\tt \:\lim_{h\to0} \: \dfrac{f(x + h) - f(x)}{h}$

$\rm :\longmapsto\:f'(x) = \displaystyle\tt \:\lim_{h\to0} \: \dfrac{ {(x + h)}^{2} - {x}^{2} }{h}$

If we substitute directly h = 0, we get indeterminant form.

$\rm :\longmapsto\:f'(x) = \displaystyle\tt \:\lim_{h\to0} \: \dfrac{ \cancel {x}^{2} + {h}^{2} + 2xh - \cancel{x}^{2} }{h}$

$\rm :\longmapsto\:f'(x) = \displaystyle\tt \:\lim_{h\to0} \: \dfrac{{h}^{2} + 2xh}{h}$

$\rm :\longmapsto\:f'(x) = \displaystyle\tt \:\lim_{h\to0} \: \dfrac{ \cancel{h}({h} + 2x)}{ \cancel{h}}$

$\rm :\longmapsto\:f'(x) = \displaystyle\tt \:\lim_{h\to0} \: (h + 2x)$

$\rm :\longmapsto\:f'(x) = 2x$

Additional Information :-

$\begin{gathered}(1)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{sin \: x}{x} \: = \: 1 }}}}}} \\ \end{gathered}$

$\begin{gathered}(2)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{tan \: x}{x} \: = \: 1 }}}}}} \\ \end{gathered}$

$\begin{gathered}(3)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{ {e}^{x} - 1 }{x} \: = \: 1 }}}}}} \\ \end{gathered}$

$\begin{gathered}(4)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{ log(1 + x) }{x} \: = \: 1 }}}}}} \\ \end{gathered}$