Math, asked by Anonymous, 1 day ago

find the second order derivative of e ^ { x } \sin 5 x




Good night guys ​

Answers

Answered by samuas980
8

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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