Question

What is the derivative of this function `e ^ sin (2x)`?

Answer 1

`d/dx (e^sin(2x)) = 2e^sin(2x)cos(2x)`

Explanation:

Using the chain rule with `y=sint` and `t= 2x` we have:

`e^sin(2x) = e^y`

`d/dx (e^sin(2x)) = d/dy (e^y) * dy/dx = e^y * dy/dx`

and similary:

`dy/dx = d/dx sint = d/dt sint * (dt)/dx = cost *dt/dx`

and finally:

`dt/dx = d/dx (2x) = 2`

Putting it together:

`d/dx (e^sin(2x)) = 2e^sin(2x)cos(2x)`

Answer 2

`2e^(sin2x)cos2x`

Explanation:

Given: `e^(sin2x)`.

Apply the chain rule, which states that,

`dy/dx=dy/(du)*(du)/dx`

Let `u=sin2x`, then we must find `(du)/dx`.

Again, use the chain rule. Let `z=2x,:.(dz)/dx=2`.

Then `y=sinz,dy/(dz)=cosz`.

Combine to get: `cosz*2=2cosz`.

Substitute back `z=2x` to get:

`=2cos(2x)`

Now, we go back to the original derivative.

From here, `y=e^u,:.dy/(du)=e^u`, and now we can combine our results from the previous findings, where `(du)/dx=2cos2x`.

`:.dy/dx=e^u*2cos2x`

`=2e^ucos2x`

Substitute back `u=sin2x` to get:

`=2e^(sin2x)cos2x`