How to find the slope of a tangent to a curve at a specific point using derivatives?

To find the slope of a tangent to a curve at a specific point using derivatives, follow these steps:

  1. Identify the point: Let's call the point where you want to find the tangent's slope as (x0,y0)(x_0, y_0). The curve is given by the function y=f(x)y = f(x).

  2. Find the derivative: Calculate the derivative of the function f(x)f(x) with respect to xx. The derivative f′(x)f'(x) gives you the slope of the tangent line at any point xx on the curve.

    For example, if f(x)=x3−3x+2f(x) = x^3 - 3x + 2, then f′(x)=ddx(x3−3x+2)=3x2−3.f'(x) = \frac{d}{dx}(x^3 - 3x + 2) = 3x^2 - 3.

  3. Evaluate the derivative at the given point: Substitute x0x_0 into the derivative function f′(x)f'(x) to find the slope of the tangent line at the point (x0,y0)(x_0, y_0).

    m=f′(x0)=3(x0)2−3m = f'(x_0) = 3(x_0)^2 - 3

  4. Write the equation of the tangent line: Now that you have the slope mm and the point (x0,y0)(x_0, y_0), you can use the point-slope form to write the equation of the tangent line at that point.

    The point-slope form is: y−y0=m(x−x0)y - y_0 = m(x - x_0)

    Substituting mm from step 3, you get: y−y0=(3(x0)2−3)(x−x0)y - y_0 = (3(x_0)^2 - 3)(x - x_0)

So, the slope of the tangent line to the curve y=f(x)y = f(x) at the point (x0,y0)(x_0, y_0) is given by f′(x0)f'(x_0), and the equation of the tangent line is: y−y0=f′(x0)(x−x0)y - y_0 = f'(x_0)(x - x_0)