# Equation of a Tangent Line (Derivative)

See how to find the equation of a tangent line

by using the derivative of a function.

2 examples and their solutions.

## Equation of a Tangent Line (Derivative)

### Formula

y = f'(a)(x - a) + f(a)

Derivative Rules

Tangent line passes through (a, f(a)).

→ y - f(a) = f'(a)(x - a)

Linear Equation (Two Variables)

→ y = f'(a)(x - a) + f(a)

### Example

f(x) = x

Tangent line at x = 1?

Solution ^{3}- 4x^{2}+ 7Tangent line at x = 1?

f(x) = x

f(1) = 1

= 1 - 4⋅1 + 7

= 1 - 4 + 7

= 4

→ (1, 4)

f'(x) = 3x

= 3x

f'(1) = 3⋅1

= 3⋅1 - 8

= 3 - 8

= -5

→ y = -5(x - 1) + 4

= -5x + 5 + 4

y = -5x + 9

^{3}- 4x^{2}+ 7f(1) = 1

^{3}- 4⋅1^{2}+ 7= 1 - 4⋅1 + 7

= 1 - 4 + 7

= 4

→ (1, 4)

f'(x) = 3x

^{2}- 4⋅2x^{1}+ 0 - [1]= 3x

^{2}- 8xf'(1) = 3⋅1

^{2}- 8⋅1= 3⋅1 - 8

= 3 - 8

= -5

→ y = -5(x - 1) + 4

= -5x + 5 + 4

y = -5x + 9

Close

### Example

f(x) = e

Tangent line at x = 0?

Solution ^{x}Tangent line at x = 0?

f(x) = e

f(0) = e

= 1 - [1]

→ (0, 1)

f'(x) = e

f'(0) = e

= 1

→ y = 1(x - 0) + 1

y = x + 1

^{x}f(0) = e

^{0}= 1 - [1]

→ (0, 1)

f'(x) = e

^{x}- [2]f'(0) = e

^{0}= 1

→ y = 1(x - 0) + 1

y = x + 1

Close