Question

find the derivation of y= tan x from first principles.
please help me...​

Answer 1

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

Given function is

$\rm :\longmapsto\:f(x) = tanx$

Change x to x + h, we get

$\rm :\longmapsto\:f(x + h) = tan(x + h)$

So, By definition of First Principal, we have

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

So, on substituting the values, we get

$\rm \:  =  \: \displaystyle\lim_{h \to 0} \frac{tan(x + h) - tanx}{h}$

$\rm \:  =  \: \displaystyle\lim_{h \to 0}\bigg[\dfrac{sin(x + h)}{cos(x + h)} - \dfrac{sinx}{cosx} \bigg]\dfrac{1}{h}$

$\rm \:  =  \: \displaystyle\lim_{h \to 0}\bigg[\dfrac{sin(x + h)cosx - sinx \: cos(x + h)}{cos(x + h) \: cosx}\bigg]\dfrac{1}{h}$

We know,

$\boxed{ \tt{ \: sinx \: cosy \: - \: siny \: cosx \: = \: sin(x - y) \: }}$

So, using this, we get

$\rm \:  =  \: \displaystyle\lim_{h \to 0}\bigg[\dfrac{sin(x + h - x)}{cos(x + h) \: cosx}\bigg]\dfrac{1}{h}$

$\rm \:  =  \: \displaystyle\lim_{h \to 0}\bigg[\dfrac{sin(h)}{cos(x + h) \: cosx}\bigg]\dfrac{1}{h}$

$\rm \:  =  \: \dfrac{1}{cosx} \times \displaystyle\lim_{h \to 0} \frac{1}{cos(x + h)} \times \displaystyle\lim_{h \to 0} \frac{sinh}{h}$

$\rm \:  =  \: \dfrac{1}{cosx} \times \dfrac{1}{cosx} \times 1$

$\rm \:  =  \: \dfrac{1}{ {cos}^{2}x}$

$\rm \:  =  \: {sec}^{2}x$

Hence,

$\rm \implies\:\boxed{ \tt{ \: \dfrac{d}{dx}tanx \: = \: {sec}^{2}x \: }}$

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Learn More :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$