Tangent from differentiation

Using differentiation, we can find the equation of the tangent to a curve at any point we like!

Finding the gradient of a tangent

The tangent is a straight line, meaning that one thing we need to find is its gradient.

As we already know, we can find the gradient of a curve at any point by differentiating the function to get its derivative (gradient function), and then substituting in the x-coordinate of the point we want the tangent at!

Find the gradient of the tangent to the curve f(x) = x^3 - 2x + 1 at the point where x = 2.

• First, we find the first derivative:
  • f'(x) = 3x^{3-1} - 2\times1x^{1-1} + 0
  • = 3x^2 - 2
• Next, we substitute in x = 2 to find the gradient of the tangent at that point:
  • f'(2) = 3(2)^2 - 2
  • = 3\times4 - 2
  • = 12 - 2
  • = 10
• Answer: The gradient of the tangent at the point where x = 2 is 10.

Finding the full equation of the tangent

Let’s take the example from above one more step and find the full equation of the tangent. Currently, we just know its gradient.

You may remember that we can find the y-intercept of any straight line if we know the gradient and one point on the line. Well, we do know a point on the line :)

We know that the tangent intersects the curve at the point where x = 2. So if we substitute x = 2 into the original function, we can find the corresponding y-coordinate, and so get the full point!

• f(2) = (2)^3 - 2(2) + 1
• = 8 - 4 + 1
• = 5
• So the tangent touches the curve at the point (2, 5).

Now we have a point, we can substitute all the information we have into the equation of a straight line to find the y-intercept, c:

• y = mx + c
• 5 = 10(2) + c
• 5 = 20 + c
• c = 5 - 20
• c = -15

Now we have our gradient and y-intercept, we can write the full equation of the tangent:

y = 10x - 15

Find the equation of the tangent to the curve f(x) = 2x^2 + 3x + 1 at the point where x = -1.

• First, we find the first derivative:
  • f'(x) = 2\times2x^{2-1} + 1\times3x^{1-1} + 0
  • = 4x + 3
• Next, we substitute in x = -1 to find the gradient of the tangent at that point:
  • f'(-1) = 4(-1) + 3
  • = -4 + 3
  • = -1
• Now, we find the corresponding y-coordinate by substituting x = -1 into the original function:
  • f(-1) = 2(-1)^2 + 3(-1) + 1
  • = 2(1) - 3 + 1
  • = 2 - 3 + 1
  • = 0
• So the tangent touches the curve at the point (-1, 0).
• Now we can substitute all the information we have into the equation of a straight line to find the y-intercept, c:
  • y = mx + c
  • 0 = -1(-1) + c
  • 0 = 1 + c
  • c = 0 - 1
  • c = -1
• Now we have our gradient and y-intercept, we can write the full equation of the tangent:
  • y = -1x - 1
  • y = -x - 1
• Answer: y = -x - 1

Example: The derivative of a curve at point (4, -2) is 5 - find the equation of the tangent

• Gradient = 5.
• X-offset = 4
• Y-offset = -2
• (y--2)=5(x-4)
• y+2=5x-20
• y=5x-22