How To Evaluate Differentials Of Functions

Differentials of Functions

for any function $f(x)$ that is differentiable,we defined $y=f(x)$ so that $\frac{dy}{dx}=f'(x)$ and the differential is given as $dy=f'(x)dx$,where the variables $dy$ and $dx$ are called the differentials of  $f(x)$

Examples
Evaluate the differentials of the following functions
(1) $f(x)=\sin{\frac{x}{2}}$ if $x=1$ and $dx=\frac{1}{2}$
(2) $f(x)=\sin2x$ if $x=\pi$ and $dx=\frac{\pi}{100}$
(3) $f(x)=\ln{x}$ if $x=1$ and $dx=\frac{1}{10}$
(4) $f(x)=2x^2-3x+5$ if $x=2$ and $dx=4$

Solution:

(1) To evaluate the differentials of $f(x)=\sin\frac{x}{2}$ we take the derivative of $f(x)$ which is
$$\begin{align*} &f'(x)=\frac{1}{2}\cos\frac{1}{2}\\ \text{use the differential formula}\\ &dy=f'(x)dx\\ &dy=\frac{1}{2}\cos\frac{x}{2}.\frac{1}{2}=\frac{1}{4}\cos\frac{x}{2}\\ &\text{substitute the initial value problem} x=1\\ &=\frac{1}{4}\cos\frac{1}{2}=\frac{1}{4}(0.999)\\ &=0.24975\end{align*}$$

(2) to evaluate the differential of  $f(x)=\sin2x$, first take the derivative of $f(x)$ and find $dy$ using the formula.
$$\begin{align*} &f(x)=\sin2x\\ &f'(x)=2\cos{2x}\\ &\text{now} dy=f'(x)dx, dy=\frac{\pi}{100}\\ &dy=2\cos{2x}.\frac{\pi}{100}=\frac{2\pi}{100}\cos{2x}\\ &\text{using the initial value problem}, x=\pi, \text{remember},\pi=180\\ &dy=\frac{\pi}{100}.2\cos{2}\pi=\frac{2\pi}{100}\\ &=\frac{\pi}{50}\\ \end{align*}$$

(3) we take the derivative of $f(x)=\ln{x}$ which is $f'(x)=\frac{1}{x}$ if $x=1$ and $dx=\frac{1}{10}$
$$\begin{align*} &dy=f'(x)dx\\ &dy=\frac{1}{x}.\frac{1}{10}\\ &\text{at}, x=1\\ &dy=\frac{1}{10}\\ \end{align*}$$

(4) the derivative of $f(x)=2x^2-3x+5$ is $f'(x)=4x-3$
$$\begin{align*} &dy=f'(x)dx\\ &dy=(4x-3).4\\ &dy=16x-12, \text{but} x=2\\ &dy=16(2)-12=32-12=20 \end{align*}$$

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