How do you find the gradient of y = x^3 at x = 2?

To find the gradient of \( y = x^3 \) at \( x = 2 \), you need to differentiate and substitute \( x = 2 \).

First, let's differentiate the function \( y = x^3 \). Differentiation is a process in calculus that helps us find the gradient (or slope) of a curve at any given point. The gradient function, also known as the derivative, tells us how steep the curve is at any point along it. For the function \( y = x^3 \), we use the power rule for differentiation. The power rule states that if \( y = x^n \), then the derivative \( \frac{dy}{dx} = nx^{n-1} \).

Applying the power rule to \( y = x^3 \), we get:
\[ \frac{dy}{dx} = 3x^2 \]

This new function, \( 3x^2 \), is the gradient function of \( y = x^3 \). It tells us the gradient of the curve at any point \( x \).

Next, we need to find the gradient at \( x = 2 \). To do this, we substitute \( x = 2 \) into the gradient function:
\[ \frac{dy}{dx} \bigg|_{x=2} = 3(2)^2 \]

Calculating this gives:
\[ 3(2)^2 = 3 \times 4 = 12 \]

So, the gradient of the curve \( y = x^3 \) at the point where \( x = 2 \) is 12. This means that at \( x = 2 \), the curve is rising steeply with a gradient of 12.