Approximations and Errors using Derivatives: Definition and Examples

Approximation Using Derivatives – Concept and Definition

Approximation in calculus helps estimate the value of a function near a given point without computing it exactly. It is particularly useful when the function is difficult to evaluate directly.

For a small change in $x$, say $\delta x$, the approximate value of the function $f(x)$ at $x = x_0 + \delta x$ is given by:

$ f(x_0 + \delta x) \approx f(x_0) + f'(x_0) \cdot \delta x $

This linear approximation gives an estimated value of $f(x)$ near $x_0$. The difference between the actual and approximate values depends on the size of $\delta x$.

Formula for Approximation

If $f(x)$ is a differentiable function, the approximate value of $f(x)$ at a nearby point can be written as:

$ f(x + \delta x) \approx f(x) + f'(x) \cdot \delta x $

Here:

• $f(x)$ = function value at $x$

• $f'(x)$ = derivative of the function at $x$

• $\delta x$ = small change in $x$

This formula helps us find the approximate value of $\mathrm{f}(\mathrm{x})$ for values close to $x$.

Example of Approximation Using Derivatives

To find $ \sqrt{25.5} $, let $ f(x) = \sqrt{x} $.
We choose $ x = 25 $ and $ \delta x = 0.5 $.

Using the approximation formula:

$f(x + \delta x) \approx f(x) + f'(x) \cdot \delta x $

We have $ f'(x) = \frac{1}{2\sqrt{x}} $.

So,
$> f(25.5) \approx \sqrt{25} + \frac{1}{2\sqrt{25}}(0.5) = 5 + \frac{0.5}{10} = 5.05 $

Hence, $ \sqrt{25.5} \approx 5.05 $ (actual value ≈ 5.0497).

Differentials and Their Relation to Approximation

Differentials are used to express small changes in variables. Let $ f: D \rightarrow R, D \subset R $ and $ y = f(x) $.
If $\Delta x$ denotes a small change in $x$, then the corresponding change in $y$ is:

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

We define:

• The differential of x, denoted by $dx$, as $> dx = \Delta x$

• The differential of y, denoted by $dy$, as $> dy = f'(x) , dx$

Thus,
$dy = \left( \frac{dy}{dx} \right) \Delta x $

This means that for very small changes, the actual change $\Delta y$ is approximately equal to the differential $dy$.

Approximations and Errors Using Derivatives

Let the function, $y=f(x)$, be a function of $x$
As we have derived derivatives earlier,

$\frac{d y}{d x}=\lim\limits_{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}=\lim\limits_{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$

$\Delta x$ is a small change in $x$ and the corresponding change in $y$ is $\Delta y$
As in the figure, point $Q$ moves closer to point $P$ on the curve, then dy is a good approximation of $\Delta y$.

$
\begin{aligned}
& \frac{d y}{d x}=\frac{\Delta y}{\Delta x}=\frac{f(x+\Delta x)-f(x)}{\Delta x} \\
\therefore \quad & \mathbf{f}(\mathbf{x}+\Delta \mathbf{x})=\mathbf{f}(\mathbf{x})+\Delta \mathbf{x} \cdot \frac{\mathbf{d y}}{\mathbf{d x}}
\end{aligned}
$

Understanding Errors in Approximations Using Derivatives

When we use derivatives to estimate values, small differences often arise between the true value and the approximated value. These small differences are called errors. Understanding the types of errors - absolute, relative, and percentage — is essential when applying differential approximations in real-life measurements, scientific calculations, or engineering problems.

1. Absolute Error – Measuring the Total Deviation

The absolute error represents the actual deviation between the true value and the measured or approximated value.

Mathematically, the absolute error in x is given by:

$ \Delta x $ or $ dx $

It tells us how much a measured quantity deviates from its true value. In practical terms, if a machine part is supposed to be 10 cm long but measures 10.02 cm, the absolute error is simply the difference — 0.02 cm.

Key takeaway: Absolute error gives the magnitude of error, without considering how large or small the true value is.

2. Relative Error – Comparing the Error to the True Value

The relative error shows how significant the absolute error is compared to the actual measurement. It expresses the error as a fraction of the true value, helping us understand the accuracy level of a measurement or approximation.

The formula for relative error in $x$ is:

$ \frac{\Delta x}{x} $ or $ \frac{dx}{x} $

This means if your absolute error is small, but your actual measurement is even smaller, the relative error could still be quite large — indicating lower precision.

In essence: Relative error is a dimensionless ratio that helps compare errors across different scales or magnitudes.

3. Percentage Error – Expressing Error in Percentage Form

The percentage error is simply the relative error expressed as a percentage, making it easier to interpret in real-world contexts.

It is given by:

$ \frac{\Delta x}{x} \times 100 $ or $ \frac{dx}{x} \times 100 $

For example, if a value of $x = 50$ has an absolute error of $0.5$, then the percentage error is:

$ \frac{0.5}{50} \times 100 = 1% $

This form is commonly used in physics experiments, engineering measurements, and quality control to indicate the accuracy of results in a clear and understandable way.

Solved Examples Based on Condition of Approximation:

Example 1: Let $f(1)=-2$ and $f^{\prime}(x) \geqslant 4.2$ for $1 \leq x \leq 6$ The possible value of $f(6)$ lies in the interval:
1) $[1,2\}$
2) $[13, \infty)$
3) $[14, \infty)$
4) $[19, \infty)$

Solution:

Given $f(1)=-2$ and $f^{\prime}(x) \geqslant 4.2$ for $1 \leq x \leq 6$
Consider $f^{\prime}(x)=\frac{f(x+h)-f(x)}{h}$

$
\Rightarrow f(x+h)-f(x)=f^{\prime}(x) \cdot h \geq(4.2) h
$
So, $f(x+h) \geq f(x)+(4.2) h$
put $x=1$ and $h=5$,
we get

$
\begin{aligned}
& f(6) \geq f(1)+5(4.2) \\
& \Rightarrow f(6) \geq 19
\end{aligned}
$
Hence $f(6)$ lies in $[19, \infty)$
Hence, the answer is option 4.

Example 2: Approximate value of $\sqrt{400.1}$ will be
1) 20.0025
2) 20.01
3) 20.001
4) 20.0035

Solution

Let $y=f(x)=\sqrt{x}$
With a small change $\Delta x$ in x , there will be a small change in y i.e., $\Delta y$

Now,

$
\begin{aligned}
& y=\sqrt{x} \Rightarrow \frac{d y}{d x}=\frac{1}{2 \sqrt{x}} \Rightarrow d y=\frac{d x}{2 \sqrt{x}} \\
& \Rightarrow \Delta y=\frac{\Delta x}{2 \sqrt{x}}\qquad . . . (i)
\end{aligned}
$
For a given question, let $\mathrm{x}=400, x+\Delta x=400.1 \Rightarrow \Delta x=0.1$
Also $\mathrm{y}=\sqrt{400}=20$

$
\begin{aligned}
& \therefore \Delta y=\frac{0.1}{2 \sqrt{400}}=0.1 / 40=0.0025 (Using (i))\\
& \therefore y+\Delta y=20+0.0025=20.0025
\end{aligned}
$

Hence, the answer is the option 1.

Example 3: Approximate value of $(999)^{\frac{1}{3}}$ equals
1) 9.96
2) 9.99
3) 10
4) 9.997

Solution

Let $\mathrm{y}=\mathrm{f}(\mathrm{x})=x^{1 / 3}$
With a small change $\Delta x$ in x , there will be a small change $\Delta y$ in y
Now,

$
\begin{aligned}
& y=x^{1 / 3} \Rightarrow \frac{d y}{d x}=\frac{1}{3 x^{2 / 3}} \\
& \frac{d y}{d x}=\frac{1}{3 x^{2 / 3}} \Rightarrow \Delta y=\frac{\Delta x}{3 x^{2 / 3}}
\end{aligned}
$
For the given question, $\mathrm{x}=1000, x+\Delta x=999 \Rightarrow \Delta x=-1$
Also $y=(1000)^{\frac{1}{3}}=10$
Now using (i)

$
\begin{aligned}
& \therefore \Delta y=\frac{-1}{3 *(1000)^{\frac{2}{3}}}=-0.00333 \\
& \therefore y+\Delta y=10-0.0033=9.99667 \approx 9.997
\end{aligned}
$
Hence, the answer is the option 4.

List of topics related to Approximations and Errors using Derivatives

This section gives you a complete overview of key subtopics connected to approximations and errors using derivatives, helping you understand where these concepts apply in calculus and real-life problem solving.

Differentiability and Existence of Derivative

Examining differentiability Using Graph of Function

Continuity of Composite Function

Continuity And Differentiability

NCERT Resources

This section compiles all NCERT study materials-including notes, textbook solutions, and exemplar exercises-to help you build a strong foundation for board exams and entrance tests.

NCERT Class 12 Maths Notes for Chapter 5 - Continuity and Differentiability

NCERT Class 12 Maths Solutions for Chapter 5 - Continuity and Differentiability

NCERT Class 12 Maths Exemplar Solutions for Chapter 5 - Continuity and Differentiability

Practice Questions based on Approximations and Errors using Derivatives

Test your understanding with targeted practice questions and MCQs on approximations and errors, designed to mirror exam-style problem-solving scenarios.

Approximations And Errors Using Derivatives- Practice Question MCQ

We have shared below the links to practice questions on the related topics to Approximations and Errors using Derivatives: