Math, asked by 12ahujagitansh, 11 hours ago

Please solve the question

f(x) =  |cosx - sinx|  \: at \: x \:  =  \:  \frac{\pi}{6}
Differentiate the given function at indicated point.​

Answers

Answered by mathdude500
5

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

Given function is

\rm :\longmapsto\:f(x) =  |cosx - sinx|

We know, By Definition of Modulus function,

\begin{gathered}\begin{gathered}\bf\:  |x|  = \begin{cases} &\sf{ - x \:  \: when \: x \:  <  \: 0} \\  \\ &\sf{ \:  \: x \:  \: when \: x \: \geqslant   \: 0} \end{cases}\end{gathered}\end{gathered}

So, given function is defined as

\begin{gathered}\begin{gathered}\bf\:  |cosx - sinx|  = \begin{cases} &\sf{ - (cosx - sinx) \:  \: when \: x \:  \geqslant   \: \dfrac{\pi}{4}} \\  \\ &\sf{ \:  \: cosx - sinx \:  \: when \: x \:  <    \: \dfrac{\pi}{4}} \end{cases}\end{gathered}\end{gathered}

So,

 \red{\rm :\longmapsto\:When \: x \:  =  \: \dfrac{\pi}{6}}

then,

\rm :\longmapsto\:f(x) = cosx - sinx

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

\rm :\longmapsto\:\dfrac{d}{dx}f(x) = \dfrac{d}{dx}( cosx -  sinx)

\rm :\longmapsto\:f'(x) =  - sinx  -  cosx

\rm :\longmapsto\:f'(x) =  - (sinx + cosx)

Thus,

\rm :\longmapsto\:f'(x)_{at \: x = \dfrac{\pi}{6}} =  -\bigg (sin\dfrac{\pi}{6} + cos\dfrac{\pi}{6}\bigg)

\rm :\longmapsto\:f'(x)_{at \: x = \dfrac{\pi}{6}} =  -\bigg (\dfrac{ \sqrt{3} }{2}  + \dfrac{1}{2} \bigg)

\rm :\longmapsto\:f'(x)_{at \: x = \dfrac{\pi}{6}} =  -\bigg (\dfrac{ \sqrt{3} + 1 }{2} \bigg)

Hence,

 \red{\rm\implies \:\boxed{\sf{ \:f'(x)_{at \: x = \dfrac{\pi}{6}}  \: =  \:  - \: \bigg (\dfrac{ \sqrt{3} + 1 }{2} \bigg)}}}

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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}

Answered by EmperorSoul
0

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

Given function is

\rm :\longmapsto\:f(x) =  |cosx - sinx|

We know, By Definition of Modulus function,

\begin{gathered}\begin{gathered}\bf\:  |x|  = \begin{cases} &\sf{ - x \:  \: when \: x \:  <  \: 0} \\  \\ &\sf{ \:  \: x \:  \: when \: x \: \geqslant   \: 0} \end{cases}\end{gathered}\end{gathered}

So, given function is defined as

\begin{gathered}\begin{gathered}\bf\:  |cosx - sinx|  = \begin{cases} &\sf{ - (cosx - sinx) \:  \: when \: x \:  \geqslant   \: \dfrac{\pi}{4}} \\  \\ &\sf{ \:  \: cosx - sinx \:  \: when \: x \:  <    \: \dfrac{\pi}{4}} \end{cases}\end{gathered}\end{gathered}

So,

 \red{\rm :\longmapsto\:When \: x \:  =  \: \dfrac{\pi}{6}}

then,

\rm :\longmapsto\:f(x) = cosx - sinx

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

\rm :\longmapsto\:\dfrac{d}{dx}f(x) = \dfrac{d}{dx}( cosx -  sinx)

\rm :\longmapsto\:f'(x) =  - sinx  -  cosx

\rm :\longmapsto\:f'(x) =  - (sinx + cosx)

Thus,

\rm :\longmapsto\:f'(x)_{at \: x = \dfrac{\pi}{6}} =  -\bigg (sin\dfrac{\pi}{6} + cos\dfrac{\pi}{6}\bigg)

\rm :\longmapsto\:f'(x)_{at \: x = \dfrac{\pi}{6}} =  -\bigg (\dfrac{ \sqrt{3} }{2}  + \dfrac{1}{2} \bigg)

\rm :\longmapsto\:f'(x)_{at \: x = \dfrac{\pi}{6}} =  -\bigg (\dfrac{ \sqrt{3} + 1 }{2} \bigg)

Hence,

 \red{\rm\implies \:\boxed{\sf{ \:f'(x)_{at \: x = \dfrac{\pi}{6}}  \: =  \:  - \: \bigg (\dfrac{ \sqrt{3} + 1 }{2} \bigg)}}}

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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}

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