Question #a1599

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Answer 1 · 2017-06-21T08:27:38

The rule for differentiating partial functions $y(x)=f(x)/g(x)$

$y'(x)=dy/dx=(df/dx*g(x)-f*dg/dx)/(g(x)^2)$

Hence

$dy/dx=(dsinx/dx*e^x-sinx*de^x/dx)/((e^x)^2)$

$dy/dx=(cosx*e^x-sinx*e^x)/(e^(2x))$

$dy/dx=[e^x*(cosx-sinx)]/(e^(2x))$

$dy/dx=(cosx-sinx)/e^x$