Question

derivative by first principal f(x) = sec√x​

Answer 1

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

Given function is

$\rm \: f(x) = sec \sqrt{x}$

So,

$\rm \: f(x + h) = sec \sqrt{x + h}$

Now, Using Definition of First Principal, we have

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

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

$\rm \: f'(x) = \displaystyle\lim_{h \to 0}\sf \: \frac{1}{h}\bigg(\dfrac{1}{cos \sqrt{x + h} } - \dfrac{1}{cos \sqrt{x} } \bigg)$

$\rm \: f'(x) = \displaystyle\lim_{h \to 0}\sf \: \frac{1}{h}\bigg(\dfrac{cos \sqrt{x} - cos \sqrt{x + h} }{cos \sqrt{x + h} \: \: cos \sqrt{x} } \bigg)$

$\rm \: f'(x) =\dfrac{1}{ {cos}^{2} \sqrt{x} } \displaystyle\lim_{h \to 0}\sf \frac{cos \sqrt{x} - cos \sqrt{x + h} }{h}$

We know,

$\boxed{\rm{  \:cosx - cosy = 2sin\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{y - x}{2} \bigg] \: }}$

So, using this result, we get

$\rm \: f'(x) =\dfrac{1}{ {cos}^{2} \sqrt{x} } \displaystyle\lim_{h \to 0}\sf \frac{2sin\bigg[\dfrac{ \sqrt{x} + \sqrt{x + h} }{2} \bigg]sin\bigg[\dfrac{ \sqrt{x + h} - \sqrt{x} }{2} \bigg]}{h}$

$\rm \: f'(x) =\dfrac{2sin \sqrt{x} }{ {cos}^{2} \sqrt{x} } \displaystyle\lim_{h \to 0}\sf \frac{sin\bigg[\dfrac{ \sqrt{x + h} - \sqrt{x} }{2} \bigg]}{h}$

$\rm \: f'(x) =\dfrac{2sin \sqrt{x} }{ {cos}^{2} \sqrt{x} } \displaystyle\lim_{h \to 0}\sf \frac{sin\bigg[\dfrac{ \sqrt{x + h} - \sqrt{x} }{2} \bigg]}{ \dfrac{ \sqrt{x + h} - \sqrt{x} }{2} } \times \frac{\dfrac{ \sqrt{x + h} - \sqrt{x} }{2} }{h}$

We know,

$\boxed{\rm{  \:\displaystyle\lim_{x \to 0}\sf \frac{sinx}{x} = 1 \: }}$

So, using this result, we get

$\rm \: f'(x) =\dfrac{sin \sqrt{x} }{ {cos}^{2} \sqrt{x} } \displaystyle\lim_{h \to 0}\sf \: 1 \times \frac{\sqrt{x + h} - \sqrt{x}}{h}$

$\rm \: f'(x) =\dfrac{sin \sqrt{x} }{ {cos}^{2} \sqrt{x} } \displaystyle\lim_{h \to 0}\sf \frac{\sqrt{x + h} - \sqrt{x}}{h} \times \dfrac{ \sqrt{x + h} + \sqrt{x} }{ \sqrt{x + h} + \sqrt{x} }$

$\rm \: f'(x) =\dfrac{sin \sqrt{x} }{ {cos}^{2} \sqrt{x} } \displaystyle\lim_{h \to 0}\sf \dfrac{ x + h - x }{h(\sqrt{x + h} + \sqrt{x}) }$

$\rm \: f'(x) =\dfrac{sin \sqrt{x} }{ {cos}^{2} \sqrt{x} } \displaystyle\lim_{h \to 0}\sf \dfrac{ h }{h(\sqrt{x + h} + \sqrt{x}) }$

$\rm \: f'(x) =\dfrac{sin \sqrt{x} }{ {cos}^{2} \sqrt{x} } \displaystyle\lim_{h \to 0}\sf \dfrac{1}{\sqrt{x + h} + \sqrt{x} }$

$\rm \: f'(x) =\dfrac{sin \sqrt{x} }{ {cos}^{2} \sqrt{x} } \times \dfrac{1}{\sqrt{x} + \sqrt{x} }$

$\rm \: f'(x) =\dfrac{sin \sqrt{x} }{ {cos}^{2} \sqrt{x} } \times \dfrac{1}{2\sqrt{x} }$

$\rm \: f'(x) =\dfrac{sin \sqrt{x} }{ {cos} \sqrt{x} \: \: cos \sqrt{x} } \times \dfrac{1}{2\sqrt{x} }$

$\rm \: f'(x) = \dfrac{sec \sqrt{x} \: tan \sqrt{x} }{2\sqrt{x} }$

$\rm\implies \:\rm \: \boxed{\rm{  \:f'(x) = \dfrac{sec \sqrt{x} \: tan \sqrt{x} }{2\sqrt{x} } \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

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