Question

If y = log (sec x), then value of y1 is​

Answer 1

Answer:

y1 = y = log (sec x)

Step-by-step explanation: y1 = y * 1 = y

So, it is the same as the question

Answer 2

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

Given function is

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

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

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

We know,

$\red{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx}f[g(x)] = f'[g(x)]\dfrac{d}{dx}g(x) \: }}}$

and

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

So, using this, we get

$\rm :\longmapsto\: y_{1} = \dfrac{1}{secx}\dfrac{d}{dx}secx$

$\rm :\longmapsto\: y_{1} = \dfrac{1}{secx} \times secx \: tanx$

$\rm \implies\:\: y_{1} \: = \: tanx$

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More to know

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