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Tangent from differentiation

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

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 at the point where .

Section titled “Find the gradient of the tangent to the curve at the point where .”
  • First, we find the first derivative:
  • Next, we substitute in to find the gradient of the tangent at that point:
  • Answer: The gradient of the tangent at the point where is .

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 . So if we substitute into the original function, we can find the corresponding y-coordinate, and so get the full point!

  • 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, :

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

Find the equation of the tangent to the curve at the point where .

Section titled “Find the equation of the tangent to the curve at the point where .”
  • First, we find the first derivative:
  • Next, we substitute in to find the gradient of the tangent at that point:
  • Now, we find the corresponding y-coordinate by substituting into the original function:
  • 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, :
  • Now we have our gradient and y-intercept, we can write the full equation of the tangent:
  • Answer: