Hello Ravi kumar,

I assume that you have asked for the derivative of the function e^x^1.22.e^sin(x).So the solution for this problem goes like this starting from this formula:

             [f(x).g(x)]'=f'(x).g(x) + f(x)g'(x)

let us assume f(x)=e^x^1.22 andf g(x)=e^sin(x)

So derivative of f(x): f'(x)=1.22x^0.22.e^x^1.22

derivative of g(x): g'(x)=x^cos(x).e^sin(x).(x^cos(x))'

 Now let us find derivative of x^cos(x):

let y=x^cos(x)

  Take log on both sides: logy= cos(x)logx

Differntiate:

1/y*y'=-sin(x)logx + cos(x)/x

y' = cos(x)/x - sin(x)logx

Now summing up: (e^x^1.22)(cos(x)/x - sin(x)logx).e^x^sin(x) + (1.22.x^0.22.e^x^1.22).e^x^sin(x)

I hope this was useful!!!

Hello

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