# Approximations of functions

## Approximations of functions

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,
$\sqrt[3]{8.06} \approx 2 + 0.01 = 2.01$.

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