Question

Find g'(x) given that g(x) = x^3 cos(πx)​

Answer 1

`3x^2 cos (\pi x)-\pi x ^3sin(\pi x)`

Explanation:

Product rule, noticing that the second function is composed.
`g'(x) = (3x^2)\cdot cos(\pi x) + (x^3) \cdot [-\pi sin(\pi x)] = 3x^2 cos (\pi x)-\pi x ^3sin(\pi x)`

Answer 2

Hello!

We can use the product rule to differentiate this function.

Let g(x) = f(x) · h(x).

Recall the following:

`(dy)/(dx) f(x) * h(x) = f'(x)h(x) + h'(x)f(x)`

In this instance, let:

`f(x) = x^3\\\\h(x) = cos(\pi x)`

Differentiate each.

f(x):

Use the product rule:

`(dy)/(dx)x^n = nx^(n-1)\\\\f'(x) = 3x^(3-1) = 3x^2`

h(x):
Use the chain rule.

`h'(x) = \pi (1) * -sin(\pi x) = -\pi sin(\pi x)`

Now, we can use the above product rule.

`g'(x) = 3x^2 * cos(\pi x) + (-\pi sin(\pi x) * x^3)\\\\\boxed{g'(x) = 3x^2cos(\pi x) - \pi x^3 sin(\pi x)}`