Question

How do you differentiate `f(x)=e^x*sin3x` using the product rule?

Answer 1

The answer is `=e^x(sin(3x)+3cos(3x))`

Explanation:

The product rule is

`(uv)'=u'v+uv'`

Here,

`f(x)=e^xsin(3x)`

and

`u=e^x`, `=>`, `u'=e^x`

`v=sin(3x)`, `=>`, `v'=3cos(3x)`

Therefore,

`f'(x)=e^xsin(3x)+3e^xcos(3x)`

`=e^x(sin(3x)+3cos(3x))`

Answer 2

`e^x(sin 3x + 3cos 3x)`

Explanation:

We know,

If `f(x) = g(x) * h(x)`,

Then, `f'(x) = d/dxf(x) = d/dx(g(x) * h(x)) = d/dxg(x) * h(x) + g(x) * d/dxh(x)`.

This is the Product Rule for Differentiation.

Now,

`f'(x)`

`= d/dxf(x)`

`= d/dx(e^x * sin 3x)`

`= d/dxe^x * sin 3x + d/dx(sin 3x) * e^x`

`= e^x sin 3x + e^x cos 3x * d/dx(3x)` [Chain Rule]

`= e^x(sin 3x + 3cos 3x)`

Hope this helps.