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

2 Answers
Aug 5, 2014

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)

Jun 14, 2015

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.