find the second order derivative of e ^ { x } \sin 5 x
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Answer:
Ur answer. dear ❤❤
Step-by-step explanation:
Let y=e
x sin5x
⇒ dxdy
= dxd (e x sin5x)=sin5x. dxd (e x )+e x dxd (sin5x) =e x (sin5x+5cos5x)
∴ dx 2 d 2 y
= dxd [ex (sin5x+5cos5x)]
=(sin5x+5cos5x) dxd (e x )+e x . dxd (sin5x+5cos5x)
=(sin5x+5cos5x)e x +e x [cos5x dxd (5x)+5(−sin5x). dxd (5x)]
=e x (sin5x+5cos5x)+e x (5cos5x−25sin5x)
=e x (10cos5x−24sin5x). =2e x (5cos5x−12sin5x)
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