Differentiation Example

Find the equation of the tangent at the point P(3,-1) when:

f(x) = x - 5x + 2x - 5

Step 1: Differentiate f(x) to find the gradient function, which is f`(x)

NBs: f`(x) is the same as dy/dx

When we differentiate we times by the power and then take 1 off the power

x ____-> (1 4)x( - ) = 4x

5x ____-> (5 3)x( - ) = 15x

2x ____-> (2 2)x( - ) = 4x = 4x

NB: Anything to the power of 1 is itself

5 = 5x ____-> (5 0)x( - ) = 0

NB: Anything to the power of 0 equals 1

Therefore f`(x) = 4x - 15x + 4x

Step 2: Sub the x coordinate for P into f`(x) - this obtains the gradient of the tangent

x = 3

f`(x) = (4 3 ) - (15 3 ) + (4 3) =

(4 27) - (15 9) + (12) =

108 - 135 + 12 = -15 = tangent gradient

NB: The gradient of the normal is the negative reciprocal of the tangent gradient (flip the number and put a negative sign in front of it) - it would be 1/15 in this case

Step 3: Sub in the x and y coordinates and the tangent gradient into:

y - y1 = mt (x - x1)

y1 = y coordinate = -1

x1 = x coordinate = 3

mt = gradient of the tangent = -15

y __1 = -15(x-3)

y + 1 = -15x + 45

The answer is: y = -15x + 44