A small change in x causes a small change in y of about the gradient times the change in x. Written out, that is δy ≈ (dy/dx) × δx, where δ means a small change.
This lesson is part of SPM Additional Mathematics differentiation. It uses the rules from differentiating powers, products and quotients.
Why does it work?
Over a very short distance, a smooth curve looks almost like its tangent. The gradient of the tangent is dy/dx, so moving δx along the curve changes y by about (dy/dx) × δx.
The smaller δx is, the better the estimate.
Worked example 1: a change in y
For y = x², find the approximate change in y when x increases from 5 to 5.02.
dy/dx = 2x = 10 at x = 5. Here δx = 0.02.
δy ≈ 10 × 0.02 = 0.2. Exactly, 5.02² − 25 = 25.2004 − 25 = 0.2004, so the estimate is close.
Worked example 2: approximating a root
Use differentiation to approximate √16.3.
Let y = √x = x^(1/2), so dy/dx = 1 ÷ (2√x). Choose x = 16, because √16 = 4 is exact, and δx = 0.3.
At x = 16: dy/dx = 1 ÷ 8. So δy ≈ (1/8)(0.3) = 0.0375.
Then √16.3 ≈ 4 + 0.0375 = 4.0375. A calculator gives 4.03733, so the approximation agrees to two decimal places (4.04) and is out by less than 0.0002.
Worked example 3: an area
The radius of a circle is 10 cm and increases by 0.2 cm. Find the approximate increase in area.
A = πr², so dA/dr = 2πr = 20π at r = 10. With δr = 0.2, δA ≈ 20π × 0.2 = 4π cm², about 12.57 cm².
The mistake that costs marks
The common slip is to start from the wrong x, or to use the calculator for the final number.
| Step | Wrong | Right |
|---|---|---|
| Start value for √16.3 | x = 16.3 | x = 16 and δx = 0.3 |
| Final answer | Calculator √16.3 = 4.0373 | 4 + 0.0375 = 4.0375, with the working shown |
Always choose the nearest value that makes the root or power exact, then say what δx is.
Check yourself
Use differentiation to approximate the cube root of 27.3.
Answer
Let y = x^(1/3), so dy/dx = (1/3)x^(−2/3). At x = 27: x^(−2/3) = 1/9, so dy/dx = 1/27.
With δx = 0.3: δy ≈ 0.3 ÷ 27 = 0.0111.
The cube root of 27.3 is approximately 3 + 0.0111 = 3.0111.
What to study next
Test the whole chapter with the differentiation practice set. If the optimisation and related-rate ideas need another look, go back to solving optimisation and rates-of-change problems.
If you would like a teacher to check your choice of starting value on new roots, see online one-to-one Additional Mathematics tuition.