we know that
$f(x + \Delta x) - f(x) =\Delta y$
$\Delta y =f(x + \Delta x) - f(x)$ -(1)
Now We define below in Approximations
The differential of x, denoted by dx, is defined by $dx = \Delta x$ -(2)
Also
$y= f(x)$
$\frac {dy}{dx} = f'(x)$
$dy= f'(x) dx$
Now from (2)
$dy= f'(x) \Delta x$
In case $dx = \Delta x$ is relatively small when compared with x, dy is a good approximation of $\Delta y$ and we denote it by $dy \approx \Delta y$
Therefore
$\Delta y = \frac {dy}{dx} \Delta x$

Solved Examples

Question 1
Find the approximate value using differentails of \(\sqrt{16.1}\). Solution
To approximate \(\sqrt{16.1}\), we'll use the function \( y =f(x) = \sqrt{x}\) near \(x = 16\).
Now
\(f'(x) = \frac{1}{2\sqrt{x}}\)
For \(x = 16\), \(f'(16) = \frac{1}{8}\).
$\Delta y = f'(16) \Delta x$
or
$\Delta y= \frac{1}{8} (0.1) = 0.0125 $
Therefore,
\(\sqrt{16.1} \approx \sqrt{16} + 0.0125 = 4.0125\).

Question 2
If \( y=f(x) = x^3 \), use the differential to approximate the change in \( f \) when \( x \) changes from 2 to 2.05 Solution
\(f(x) = x^3\).
Now
\(f'(x) = 3x^2\).
For \(x = 2\), \(f'(2) = 12\).
$\Delta y = f'(2) \Delta x$
\[ \Delta y \approx 12(0.05) = 0.6 \]

Question 3
Approximate the cube root of 8.06 using differentials? Solution
\( y= f(x) = x^{\frac{1}{3}}\):
Now
\(f'(x) = \frac{1}{3x^{\frac{2}{3}}}\).
For \(x = 8\), \(f'(8) = \frac{1}{6}\).
$\Delta y = f'(8) \Delta x$
\[ \Delta y \approx \frac{1}{6} (0.06) = 0.01 \]
Therefore,
\(\sqrt[3]{8.06} \approx 2 + 0.01 = 2.01\).