Let *f* be a function defined on to and C be the curve whose equation is y = *f *(x). This article shows us how to determine the equation of the tangent line to C at the point K(*a*,*b*) where *b* = *f *(*a*). Look at the figure below.

**Figure 1**

The equation of the tangent line is determined by the following formula:

…………………………………………………………………………………………………………………………………………………………………… (*)

where *f* ‘(*a*) is the derivative of *f* at x = a.

**Example 1**

Find the equation of the tangent line to the curve given by y = x^{3} – 2x^{2} at the point K(1,-1).

**Answer**

In this example, so . Substituting x = 1 (i.e. the abscissa of K) into , we get . By substituting a = 1, b = -1, and into (*), we have:

y – (-1) = (-1)⋅(x – 1)

y + 1 = -x + 1

y = -x

So, the equation of the tangent line to the curve at K is y = -x. (See Figure 2.)

**Figure 2**

**Example 2**

Find the equation of the tangent line to the curve with equation y = 3x^{5} – 5x^{3} at point K(-1,2).

**Answer**

In this example, so . Substituting x = -1 (i.e. the abscissa of K) into , we get . By substituting a = -1 , b = 2, and into (*), we get:

y – 2 = 0(x – (-1))

y – 2 = 0

y = 2

So, the equation of the tangent line to the curve at K is y = 2. (See Figure 3.)

**Figure 3**

**Example 3**

Find the equation of the tangent to the curve with equation at x = 4.

**Answer**

In this example, . Then . Substituting x = 4 into the , we get . To determine the ordinate of the point whose abscissa is 4, substitute x = 4 into . This gives y = 4. The equation of the tangent line to the curve at the point (4,4) will be obtained by substituting a = 4, b = 4, and into (*). This yields:

Thus, the equation of the tangent line to the curve at x = 4 is . (See Figure 4.)

**Figure 4**

**Example 4**

Find the equation of the tangent line to the curve with the equation y = 4 sin x at the point whose abscissa is π/3.

**Answer**

In this example, f(x) = 4 sin x so f ‘(x) = 4 cos x. Substituting x = π/3 into the f ‘(x), we have f ‘(π/3) = 4 cos π/3 = 4⋅½ = 2. To determine the ordinate of the point whose abscissa is π/3, substitute x = π/3 into y = 4 sin x. This gives . The equation of the tangent line to the curve at the point will be obtained by substituting a = π/3, b = 2√3, and f ‘(π/3) = 2 into (*). This results in:

y – 2√3 = 2(x – π/3)

y – 2√3 = 2x – 2π/3

y = 2x + (2√3 – 2π/3)

So, the equation of the tangent line to the curve at x = π/3 is y = 2x + (2√3 – 2π/3). (See Figure 5.)

**Figure 5**