Question

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

Answer 1

Answer:

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

Step-by-step 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:
$\frac{dy}{dx} f(x) * h(x) = f'(x)h(x) + h'(x)f(x)$

In this instance, let:
$f(x) = x^3\\\\h(x) = cos(\pi x)$

Differentiate each.

f(x):

Use the product rule:
$\frac{dy}{dx}x^n = nx^{n-1}\\\\f'(x) = 3x^{3-1} = 3x^2$

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.

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