Question

If y=log (secx+tanx)then dy/dx is equal to

Answer 1

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

Given function is

$\rm :\longmapsto\:y = log[secx + tanx]$

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

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

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx}logx = \frac{1}{x} \: \: }}$

So, using this, we get

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

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx}(u + v) = \dfrac{d}{dx} u+ \dfrac{d}{dx}v \: }}$

So, using this, we get

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

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx}tanx = {sec}^{2}x}} \: \: and \: \: \boxed{ \tt{ \: \dfrac{d}{dx}secx = secxtanx}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{secx + tanx} \: \bigg[secx \: tanx + {sec}^{2}x \bigg]$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{secx + tanx} \: secx\bigg[ \: tanx + {sec}^{}x \: \bigg]$

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

More to know :-

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

Answer 2

ANSWER:

Given:

• y = log(sec x + tan x)

To Find:

• dy/dx

Solution:

We are given that,

$\implies y=\log(\sec x+\tan x)$

Taking derivative of y,

$\implies\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg(\log(\sec x+\tan x)\bigg)$

We know that, by Chain Rule,

$\hookrightarrow[f(g(x))]'=f'g(x)\times g'(x)$

So, on applying Chain Rule,

$\implies\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg(\log(\sec x+\tan x)\bigg)$

$\implies\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg(\log(\sec x+\tan x)\bigg)\times\dfrac{d}{dx}\bigg(\sec x+\tan x\bigg)\times\dfrac{d}{dx}\bigg(x\bigg)$

$\implies\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg(\log(\sec x+\tan x)\bigg)\times\left(\dfrac{d}{dx}(\sec x)+\dfrac{d}{dx}(\tan x)\right)\times\dfrac{d}{dx}\bigg(x\bigg)$

We know that,

$\hookrightarrow\dfrac{d}{dx}\log(x)=\dfrac{1}{x}$

$\hookrightarrow\dfrac{d}{dx}\sec x=\sec x \tan x$

$\hookrightarrow\dfrac{d}{dx}\tan x=\sec^2x$

$\hookrightarrow\dfrac{d}{dx} x=1$

So,

$\implies\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg(\log(\sec x+\tan x)\bigg)\times\left(\dfrac{d}{dx}(\sec x)+\dfrac{d}{dx}(\tan x)\right)\times\dfrac{d}{dx}\bigg(x\bigg)$

$\implies\dfrac{dy}{dx}=\dfrac{1}{\sec x+\tan x}\times\left(\sec x\tan x+\tan^2x\right)\times1$

$\implies\dfrac{dy}{dx}=\dfrac{\sec x\tan x+\sec^2x}{\sec x+\tan x}$

Taking sec x common in numerator,

$\implies\dfrac{dy}{dx}=\dfrac{\sec x(\tan x+\sec x)}{\sec x+\tan x}$

$\implies\dfrac{dy}{dx}=\dfrac{\sec x(\sec x+\tan x)}{(\sec x+\tan x)}$

Cancelling (sec x + tan x),

$\implies\dfrac{dy}{dx}=\dfrac{\sec x}{1}$

Hence,

$\implies\bf\dfrac{dy}{dx}=sec\,x$