ত্রিকোনমিতিক ফাংশনের অন্তরজ

If the prime sign (') represents differentiation w.r.t. xx and f′=sin⁡x+sin⁡4x.cos⁡xf^{'}=\sin x+\sin 4x.\cos x, then f′(2x2+π2)f^{'}\left ( 2x^{2}+\cfrac{\pi }{2} \right ) at x=π2x=\sqrt{\dfrac{\pi }{2}} is equal to

কাজু বাদাম

Given f′(x)=sin⁡x+sin⁡4x.cos⁡xf^{'}(x)=\sin x+\sin 4x.\cos x
Now, f′(2x2+π2)f^{'}\left ( 2x^{2}+\dfrac{\pi }{2} \right )
=f′(2x2+π2)ddx(2x2+π2)=f^{ ' }\left( 2x^{ 2 }+\dfrac { \pi }{ 2 } \right) \dfrac { d }{ dx } \left( 2x^{ 2 }+\dfrac { \pi }{ 2 } \right)

=[sin⁡(2x2+π2)+sin⁡4(2x2+π2).cos⁡(2x2+π2)]4x=\left[ \sin \left( 2x^{ 2 }+\dfrac { \pi }{ 2 } \right) +\sin 4\left( 2x^{ 2 }+\dfrac { \pi }{ 2 } \right) .\cos \left( 2x^{ 2 }+\dfrac { \pi }{ 2 } \right) \right] 4x

=[cos⁡2x2−sin⁡8x2.sin⁡2x2]4x=\left[ \cos 2x^{ 2 }-\sin 8x^{ 2 }.\sin 2x^{ 2 } \right] 4x

=[cos⁡π−sin⁡4π.sin⁡π]4π2=\left[ \cos \pi -\sin 4\pi .\sin \pi \right] 4\sqrt { \dfrac { \pi }{ 2 } }
=−22π=-2\sqrt{2\pi}

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