Basic Differentiation
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Example 11:
$$ y=4x^3-\frac{5}{x^2},\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$
Solution:
\begin{align*}
y&=4x^3-\frac{5}{x^2}\\
\frac{dy}{dx}&=\frac{d}{dx}\left(4x^3-\frac{5}{x^2}\right)\\
\frac{dy}{dx}&=\frac{d}{dx}4x^3-\frac{d}{dx}\frac{5}{x^2}\\
\frac{dy}{dx}&=\frac{d}{dx}4x^3-\frac{d}{dx}5x^{-2}\\
\frac{dy}{dx}&=4\frac{d}{dx}x^3-5\frac{d}{dx}x^{-2}\\\\
\because\frac{d}{dx}x^n&=nx^{n-1}\\\\
\therefore\;\; \frac{dy}{dx}&=4(3x^{3-1})-5(-2x^{-2-1})\\
\frac{dy}{dx}&=12x^2+10x^{-3}\\
\frac{dy}{dx}&=12x^2+\frac{10}{x^3}\\
\end{align*}
Example 12:
$$ y=\frac{8}{x}+\frac{6}{x^2}+\frac{7}{x^3},\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$
Solution:
\begin{align*}
y&=\frac{8}{x}+\frac{6}{x^2}+\frac{7}{x^3}\\
\frac{dy}{dx}&=\frac{d}{dx}\left(\frac{8}{x}+\frac{6}{x^2}+\frac{7}{x^3},\right)\\
\frac{dy}{dx}&=\frac{d}{dx}\frac{8}{x}+\frac{d}{dx}\frac{6}{x^2}+\frac{d}{dx}\frac{7}{x^3}\\
\frac{dy}{dx}&=\frac{d}{dx}8x^{-1}+\frac{d}{dx}6x^{-2}+\frac{d}{dx}7x^{-3}\\
\frac{dy}{dx}&=8\frac{d}{dx}x^{-1}+6\frac{d}{dx}x^{-2}+7\frac{d}{dx}x^{-3}\\\\
\because\frac{d}{dx}x^n&=nx^{n-1}\\\\
\therefore\;\; \frac{dy}{dx}&=8(-1x^{-1-1})+6(-2x^{-2-1})+7(-3x^{-3-1})\\
\frac{dy}{dx}&=-8x^{-2}-12x^{-3}-21x^{-4}\\
\frac{dy}{dx}&=-\frac{8}{x^2}-\frac{12}{x^3}-\frac{21}{x^3}\\
\end{align*}
Example 13:
$$ y=6x^4+\frac{3}{x^5},\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$
Solution:
\begin{align*}
y&=6x^4+\frac{3}{x^5}\\
\frac{dy}{dx}&=\frac{d}{dx}\left(6x^4+\frac{3}{x^5}\right)\\
\frac{dy}{dx}&=\frac{d}{dx}6x^4+\frac{d}{dx}\frac{3}{x^5}\\
\frac{dy}{dx}&=\frac{d}{dx}6x^4+\frac{d}{dx}3x^{-5}\\
\frac{dy}{dx}&=6\frac{d}{dx}x^4+3\frac{d}{dx}x^{-5}\\\\
\because\frac{d}{dx}x^n&=nx^{n-1}\\\\
\therefore\;\; \frac{dy}{dx}&=6(4x^{4-1})+3(-5x^{-5-1})\\
\frac{dy}{dx}&=24x^3-15x^{-6}\\
\frac{dy}{dx}&=24x^3-\frac{15}{x^6}\\
\end{align*}
Example 14:
$$ y=\sqrt{x},\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$
Solution:
\begin{align*}
y&=\sqrt{x}\\
\frac{dy}{dx}&=\frac{d}{dx} \sqrt{x}\\
\frac{dy}{dx}&=\frac{d}{dx} x^{\frac{1}{2}}\\\\
\because\;\; \frac{d}{dx} x^n&=nx^{n-1}\\\\
\therefore\;\; \frac{dy}{dx}&=\frac{1}{2}x^{\frac{1}{2}-1}\\
\frac{dy}{dx}&=\frac{1}{2}x^{\frac{1-2}{2}}\\
\frac{dy}{dx}&=\frac{1}{2}x^{-\frac{1}{2}}\\
\frac{dy}{dx}&=\frac{1}{2x^{\frac{1}{2}}}\\
\frac{dy}{dx}&=\frac{1}{2\sqrt{x}}\\
\end{align*}
Example 15:
$$ y=\frac{1}{\sqrt{x}},\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$
Solution:
\begin{align*}
y&=\frac{1}{\sqrt{x}}\\
\frac{dy}{dx}&=\frac{d}{dx} \frac{1}{\sqrt{x}}\\
\frac{dy}{dx}&=\frac{d}{dx} \frac{1}{x^{\frac{1}{2}}}\\
\frac{dy}{dx}&=\frac{d}{dx} x^{-\frac{1}{2}}\\\\
\because\;\; \frac{d}{dx} x^n&=nx^{n-1}\\\\
\therefore\;\; \frac{dy}{dx}&=-\frac{1}{2}x^{-\frac{1}{2}-1}\\
\frac{dy}{dx}&=-\frac{1}{2}x^{\frac{-1-2}{2}}\\
\frac{dy}{dx}&=-\frac{1}{2}x^{-\frac{3}{2}}\\
\frac{dy}{dx}&=-\frac{1}{2x^{\frac{3}{2}}}\\
\frac{dy}{dx}&=-\frac{1}{2x^{1+\frac{1}{2}}}\\
\frac{dy}{dx}&=-\frac{1}{2x^1x^{\frac{1}{2}}}\\
\frac{dy}{dx}&=-\frac{1}{2x\sqrt{x}}\\
\end{align*}