# Differentials Errors and Approximations

Here you will learn what is differentials errors and approximations with examples.

Let’s begin –

## Differentials Errors and Approximations

In order to calculate the approximate value of a function, differentials may be used where the differential of a function is equal to its derivative multiplied by the differential of the independent variable,

In general dy = f'(x)dx or df(x) = f'(x)dx

Or $$\Delta y$$ = $$dy\over dx$$ $$\Delta x$$

Because [ f'(x) = $$dy\over dx$$ ]

#### Absolute Error

The error $$\Delta x$$ in x is called the absolute error in x.

#### Relative Error

If $$\Delta x$$ is an error in x, then $$\Delta x\over x$$ is called the relative error in x

#### Percentage Error

If $$\Delta x$$ is an error in x, then $${\Delta x\over x} \times 100$$ is called the percentage error in x.

Remark : Let y = f(x) be a function of x, and let $$\Delta x$$ be a small change in x. Let the corresponding change in y be $$\Delta y$$. Then,

y + $$\Delta y$$ = $$f(x + \Delta x)$$

But, $$\Delta y$$ = $$dy\over dx$$ $$\Delta x$$ = f'(x) $$\Delta x$$ , approximately

$$\therefore$$ $$f(x + \Delta x)$$ = y + $$\Delta y$$

$$\implies$$ $$f(x + \Delta x)$$ = y + f'(x) $$\Delta x$$ , approximately

$$\implies$$ $$f(x + \Delta x)$$ = y + $$dy\over dx$$ $$\Delta x$$ , approximately

Let x be the independent variable and y be the dependent variable connected by the relation y = f(x). We use the following algorithm to find an approximate change $$\Delta y$$ in y due to small change $$\Delta x$$ in x.

Algorithm :

1). Choose the initial value of the independent variable as x and the changed value as x + $$\Delta x$$.

2). find $$\Delta x$$ and assume that dx = $$\Delta x$$.

3). find $$dy\over dx$$ from the given relation y = f(x).

4). find the value of $$dy\over dx$$ at (x, y).

5). find dy by using the relation dy = $$dy\over dx$$dx.

6). Put $$\Delta y$$ = dy to obtain an approximate change in y.

Example : If y = $$x^4$$ – 10  and x changes from 2 to 1.99, what is the approximate change in y ? Also, find the changed value of y.

Solution : Let x  = 2, x + $$\Delta x$$ = 1.99. Then, $$\Delta x$$ = 1.99 – 2 = -0.01

Let dx = $$\Delta x$$  = -0.01

We have,

y = $$x^4$$ – 10

$$\implies$$ $$dy\over dx$$ = $$4x^3$$ $$\implies$$ $$({dy\over dx})_{x=2}$$ = $$4(2)^3$$ = 32

$$\therefore$$ dy = $$dy\over dx$$ dx

$$\implies$$ dy = 32(-0.01) = -0.32

$$\implies$$ $$\Delta y$$ = -0.32 approximately

So, approximate change in y = -0.32

When, x = 2, we have

y = $$2^4$$ – 10 = 6

So, changed value of y = y + $$\Delta y$$ = 6 + (-0.32) = 5.68

### Related Questions

If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximating error in calculating its volume.

Find the approximate value of f(3.02), where f(x) = $$3x^2 + 5x + 3$$.