Math, asked by sajan6491, 15 hours ago

 \small\displaystyle \tt   \red{\frac{\partial}{\partial x} \bigg[7 \cdot \sin \bigg(5 \cdot x -  \frac{\pi}{2} \bigg )  +  \cos \bigg( 3 \cdot x +  \frac{\pi}{2} \bigg) \bigg ] \bigg|_{x = 2\pi}}

Answers

Answered by mathdude500
18

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

Cosider,

\rm \:  7 \cdot \sin \bigg(5 \cdot x - \dfrac{\pi}{2} \bigg ) + \cos \bigg( 3 \cdot x + \dfrac{\pi}{2} \bigg)

We know,

\boxed{\tt{ sin( - x) =  - sinx \: }} \\

So, using this result, the above expression can be rewritten as

\rm \: =  \:   -  7 \sin \bigg(\dfrac{\pi}{2} - 5x \bigg ) + \cos \bigg(\dfrac{\pi}{2} + 3x \bigg)

\rm \:  =  \:  - 7cos5x - sin3x

So,

\rm \: \dfrac{\partial}{\partial x} \bigg[7 \cdot \sin \bigg(5 \cdot x - \dfrac{\pi}{2} \bigg ) + \cos \bigg( 3 \cdot x + \dfrac{\pi}{2} \bigg) \bigg ]

can be rewritten as

\rm \:  =  \: \dfrac{\partial}{\partial x}\bigg( - 7cos5x - sin3x\bigg)

\rm \:  =  \:  - 7\dfrac{\partial}{\partial x}cos5x - \dfrac{\partial}{\partial x}sin3x

We know,

\boxed{\tt{ \dfrac{d}{dx}sinx \:  =  \: cosx \: }} \\

and

\boxed{\tt{ \dfrac{d}{dx}cosx \:  =  \:  -  \: sinx \: }} \\

So, using these results, we get

\rm \:  =  \:  - 7 \: ( - sin5x)\dfrac{d}{dx}5x - cos3x\dfrac{\partial}{\partial x}3x

\rm \:  =  \: 7sin5x \times 5 - cos3x \times 3

\rm \:  =  \: 35sin5x - 3cos3x

Hence,

\rm \: \dfrac{\partial}{\partial x} \bigg[7 \cdot \sin \bigg(5 \cdot x - \dfrac{\pi}{2} \bigg ) + \cos \bigg( 3 \cdot x + \dfrac{\pi}{2} \bigg) \bigg ]_{x = 2\pi}

\rm \:  =  \: 35sin10\pi \:  -  \: 3cos6\pi

We know,

\boxed{\tt{ sin \: n\pi = 0 \:  \:  \: where \: n \: is \: integer}} \\

and

\boxed{\tt{ cos \: n\pi =  {( - 1)}^{n} \:  \:  \: where \: n \: is \: integer}} \\

So, using these results, we get

\rm \:  =  \: 0 - 3 {( - 1)}^{6}

\rm \:  =  \:  - 3  \times 1

\rm \:  =  \:  - 3

Hence,

\boxed{\tt{ \rm \: \dfrac{\partial}{\partial x} \bigg[7 \cdot \sin \bigg(5 \cdot x - \dfrac{\pi}{2} \bigg ) + \cos \bigg( 3 \cdot x + \dfrac{\pi}{2} \bigg) \bigg ]_{x = 2\pi} =  - 3}} \\

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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}

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