Linear Approximations

One of the main uses of the derivative is linearization, which uses the first two terms (the constant and linear term) of the Taylor series as an approximation. In many applications, this gives a very good approximation, as we will see in some examples.

Linear variation visualized

There are several geometric examples where it is possible to see the linear variation as the change in area as a parameter is changed by a small amount.


The area of a square of side length is given by . When that is varied by a small amount amount , the result is

The linear variation is , which can be seen in the diagram as the rectangles along the right and top edges of the square. There are two of them, each with area . The final bit of area, the purple square in the diagram, has area , which is higher order.


The area of a right triangle with legs length is . When the leg is varied by , the result is

Visually, the linear variation comes from the red parallelogram, of base and height , running along the hypotenuse. The higher order term comes from the small purple triangle at the tip of the triangle.


The area of a disc of radius is . If the radius is increased by , the result is

Visually breaking this into the linear variation and higher order variation is a little bit harder. The best way is to imagine taking the ring formed by the increased radius and breaking it into rectangles and wedges. In the limit, the wedges can be rearranged into a disc of radius , and the rectangles can be arranged to form a strip of length (the circumference of the inner circle) and width .

Linear approximation

The equation underlying any linear approximation should be familiar, since it is just the first order Taylor series about , after making the substitution :

This will be a good linear approximation provided that is small, i.e., the point is close to the input we are trying to approximate. In general, one wants to pick to be an input where it is easy to compute and which is as close to the desired input as possible.

Example Using a linear approximation, estimate .

The function here is . Possible choices for are perfect squares, because it is easy to compute the square root of squares. The nearest perfect square is , so we choose . Thus, . Then

which is very close to the calculator's answer of .

Example Using a linear approximation, estimate . Is this an over-approximation or an under-approximation?

As above, is the closest point where it is easy to compute and derivatives. Then , so the linear approximation is

This is an over-approximation. One way to see why is to consider the graph of , which is concave down, so the linear approximation is above the true value. Another argument is to consider the next term of the Taylor series, which is negative (since the second derivative of is negative).

For comparison, the value according to a calculator is .

Example Approximate . Hint: .

From the hint, . Thus, we are trying to approximate at . The nearest easy input is , so we find

The true answer is approximately , so this estimate is within 2%.

Example Approximate . Hint: , and .

From the first hint, , so consider the linear approximation for near :

So it follows that

(the last step used the second hint that ). The true answer is approximately , so the error is less than 2%.

Newton's method

Another application of linearization gives a way of approximating the root of a function. This is called Newton's method.

Newton's method uses a difference equation to approximate a root of a continuous function. Given a continuous function , choose an initial guess of a root of the function, and then iterate the equation

The resulting sequence hopefully converges to a root of . Graphically, what is happening is as follows:

  1. Pick a guess .
  2. Find the tangent line to through the point .
  3. Let be the point where the tangent line intersects the -axis.
  4. Repeat steps 2 and 3 (see the figure).


This sequence is only defined if exists and is non-zero for every in the sequence. Even if the sequence is defined, it may not converge to anything. But if the sequence is defined and it does converge, say to , then is a root of .


Find the difference equation for using Newton's method to approximate .

The first step is to find a function whose root is . A good choice is . Then according to Newton's method, the difference equation is