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If y=sinx+e^x ,t h e n(d^(2x))/(dy^2)= ...

If `y=sinx+e^x ,t h e n(d^(2x))/(dy^2)=` `(-sinx+e^x)^(-1)` `(sinx-e^x)/((cosx+e^x)^2)` (c) `(sinx-e^x)/((cosx+e^x)^3)` (d) `(sinx+e^x)/((cosx+e^x)^3)`

A

`(-sin x +e^(x))^(-1)`

B

`(sin x - e^(x))/((cos x +e^(x))^(2))`

C

`(sin x - e^(x))/((cos x +e^(x))^(3))`

D

`(sin x + e^(x))/((cos x +e^(x))^(3))`

Text Solution

Verified by Experts

`y=sin x +e^(x)or (dy)/(dx)=cos x +e^(x)`
`"or "(dx)/(dy)=(cos x +e^(x))^(-1)" (1)"`
`therefore" "(d^(2)x)/(dy^(2))=-(cos x +e^(x))^(-2)(-sin x +e^(x))(dx)/(dy)`
Substituting the value of `(dy)/(dx)` from (1),
`(d^(2)x)/(dy^(2))=((sin x -e^(x)))/((cos x +e^(x))^(2))(cos x +e^(x))^(-1)=(sin x-e^(x))/((cos x +e^(x))^(3))`
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