Question

How do you find the derivative of `y=e^x cos(x)` ?

Answer 1

This is a type of problem involving the product rule.

The product rule states:

`d/dx[f(x) * g(x)] = f'(x)g(x) + f(x)g'(x)`

So, we will let `f(x) = e^x`, and `g(x) = cos x`.

We know that the derivative of `e^x` is simply `e^x`, and that the derivative of `cos x` is equal to `-sin x`.

(if these identities look unfamiliar to you, I may recommend viewing videos from this page or this page, which explain the derivative rules for `e^x` and `cos x` more in-depth)

Therefore, `f'(x) = e^x`, and `g'(x) = -sin x`. We can then simply substitute into the product rule formula:

`d/dx[e^x cos x] = e^x * (-sin x) + e^x * cos x`

To make this equation a little prettier, we will factor the `e^x`:

`d/dx[e^x cos x] = e^x * (cos x - sin x)`

Answer 2

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

Explanation:

When two variables are multiplied in derivative
We Have Formula,
`(d(uv))/dx=u.((d(v))/dx)+ v.((d(u))/dx)`
Question: To find the derivative of `y=e^xcos(x)` ?

Diffentiating both side by `x`
We get :
`dy/dx=(d(e^xcos(x)))/dx`

`dy/dx=e^x ((d(cosx))/dx)+cosx ((d(e^x ))/dx)`

Here,
`(d(cosx))/dx=-sinx` for first derivative.

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

Which is required solution.