Question

d/dx(e* sinx)= how much​

Answer 1

Answer:

Before finding the derivative of y=esin(x) ,I would like to recall the derivative of y=ex ---1 which is dydx=ex

Before finding the derivative of y=esin(x) ,I would like to recall the derivative of y=ex ---1 which is dydx=ex Next we need knowledge of chain rule to find out the derivative of

Before finding the derivative of y=esin(x) ,I would like to recall the derivative of y=ex ---1 which is dydx=ex Next we need knowledge of chain rule to find out the derivative ofy=esin(x) ---2

Before finding the derivative of y=esin(x) ,I would like to recall the derivative of y=ex ---1 which is dydx=ex Next we need knowledge of chain rule to find out the derivative ofy=esin(x) ---2Simply think of x in place of sin(x) in 2 and differentiate, we get esin(x) again we need to differentiate sin(x) we get cos(x) .

Before finding the derivative of y=esin(x) ,I would like to recall the derivative of y=ex ---1 which is dydx=ex Next we need knowledge of chain rule to find out the derivative ofy=esin(x) ---2Simply think of x in place of sin(x) in 2 and differentiate, we get esin(x) again we need to differentiate sin(x) we get cos(x) .It is depicted clearly as,

Before finding the derivative of y=esin(x) ,I would like to recall the derivative of y=ex ---1 which is dydx=ex Next we need knowledge of chain rule to find out the derivative ofy=esin(x) ---2Simply think of x in place of sin(x) in 2 and differentiate, we get esin(x) again we need to differentiate sin(x) we get cos(x) .It is depicted clearly as,y=esin(x)

Before finding the derivative of y=esin(x) ,I would like to recall the derivative of y=ex ---1 which is dydx=ex Next we need knowledge of chain rule to find out the derivative ofy=esin(x) ---2Simply think of x in place of sin(x) in 2 and differentiate, we get esin(x) again we need to differentiate sin(x) we get cos(x) .It is depicted clearly as,y=esin(x) =>dydx=esin(x).dsin(x)dx.dxdx

Before finding the derivative of y=esin(x) ,I would like to recall the derivative of y=ex ---1 which is dydx=ex Next we need knowledge of chain rule to find out the derivative ofy=esin(x) ---2Simply think of x in place of sin(x) in 2 and differentiate, we get esin(x) again we need to differentiate sin(x) we get cos(x) .It is depicted clearly as,y=esin(x) =>dydx=esin(x).dsin(x)dx.dxdx =>dydx=esin(x).cos(x).1

Before finding the derivative of y=esin(x) ,I would like to recall the derivative of y=ex ---1 which is dydx=ex Next we need knowledge of chain rule to find out the derivative ofy=esin(x) ---2Simply think of x in place of sin(x) in 2 and differentiate, we get esin(x) again we need to differentiate sin(x) we get cos(x) .It is depicted clearly as,y=esin(x) =>dydx=esin(x).dsin(x)dx.dxdx =>dydx=esin(x).cos(x).1 =>dydx=cos(x)esin(x)

Before finding the derivative of y=esin(x) ,I would like to recall the derivative of y=ex ---1 which is dydx=ex Next we need knowledge of chain rule to find out the derivative ofy=esin(x) ---2Simply think of x in place of sin(x) in 2 and differentiate, we get esin(x) again we need to differentiate sin(x) we get cos(x) .It is depicted clearly as,y=esin(x) =>dydx=esin(x).dsin(x)dx.dxdx =>dydx=esin(x).cos(x).1 =>dydx=cos(x)esin(x) I hope this clear your application of chain rule as well as answer.