Question

if y= log ( secx+ tanx) then find dy/dx​

Answer 1

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

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

$\boxed{ \red{\bf{\dfrac{d}{dx}tanx = {sec}^{2}x }}}$

$\boxed{ \red{\bf{\dfrac{d}{dx}secx = secxtanx}}}$

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

Given that,

$\rm :\longmapsto\:y = log(secx + tanx)$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} y = \dfrac{d}{dx} log(secx + tanx)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{secx + tanx} \dfrac{d}{dx} (secx + tanx)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{secx + tanx} \bigg(\dfrac{d}{dx} secx + \dfrac{d}{dx} tanx \bigg)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{secx + tanx} \times ( {sec}^{2}x + secx \: tanx)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{secx + tanx} \times secx( {sec}x \: + \: tanx)$

$\bf\implies \:\dfrac{dy}{dx} = secx$

Additional Information :-

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

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

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

$\boxed{ \red{\bf{\dfrac{d}{dx}sinx = cosx}}}$

$\boxed{ \red{\bf{\dfrac{d}{dx}cosx = - sinx}}}$

$\boxed{ \red{\bf{\dfrac{d}{dx}cosecx = - cosecx \: cotx}}}$

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