Basic Differentiation

Basic Differentiation

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Example 1:

$$ y=x^4,\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y&=x^4\\ \frac{dy}{dx} &=\frac{d}{dx} x^4\\\\ \because\;\; \frac{d}{dx} x^n &=nx^{n-1}\\\\ \therefore\;\; \frac{dy}{dx} &=4x^{4-1}\\ \frac{dy}{dx} &=4x^3\\ \end{align*}

Example 2:

$$ y=3x^5,\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y &=3x^5\\ \frac{dy}{dx} &=\frac{d}{dx} 3x^5\\ \frac{dy}{dx} &=3\frac{d}{dx}x^5\\\\ \because\;\; \frac{d}{dx} x^n &=nx^{n-1}\\\\ \therefore\;\; \frac{dy}{dx} &=3(5x^{5-1})\\ \frac{dy}{dx} &=15x^4\\ \end{align*}

Example 3:

$$ y=x^3+8,\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y &=x^3+8\\ \frac{dy}{dx} &=\frac{d}{dx}(x^3+8)\\ \frac{dy}{dx} &=\frac{d}{dx}x^3+\frac{d}{dx}8\\\\ \because\;\; \frac{d}{dx} x^n &=nx^{n-1}\;\; \textup{and} \;\:\frac{d}{dx} c=0\\\\ \therefore\;\; \frac{dy}{dx} &=3x^{3-1}+0\\ \frac{dy}{dx} &=3x^2\\ \end{align*}

Example 4:

$$ y=2x^2+5x,\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y&=2x^2+5x\\ \frac{dy}{dx}&=\frac{d}{dx}(2x^2+5x)\\ \frac{dy}{dx}&=\frac{d}{dx}2x^2+\frac{d}{dx}5x\\ \frac{dy}{dx}&=2\frac{d}{dx}x^2+5\frac{d}{dx}x\\\\ \because\;\; \frac{d}{dx}x^n&=nx^{n-1}\;\; \textup{and} \;\:\frac{d}{dx}x=1\\\\ \therefore\;\; \frac{dy}{dx}&=2(2x^{2-1})+5(1)\\ \frac{dy}{dx}&=4x+5\\ \end{align*}

Example 5:

$$ y=7x^6-9x^4+12x^2,\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y&=7x^6-9x^4+12x^2\\ \frac{dy}{dx}&=\frac{d}{dx}(7x^6-9x^4+12x^2)\\ \frac{dy}{dx}&=\frac{d}{dx}7x^6-\frac{d}{dx}9x^4+\frac{d}{dx}12x^2\\ \frac{dy}{dx}&=7\frac{d}{dx}x^6-9\frac{d}{dx}x^4+12\frac{d}{dx}x^2\\\\ \because\;\; \frac{d}{dx}x^n&=nx^{n-1}\\\\ \therefore\;\; \frac{dy}{dx}&=7(6x^{6-1})-9(4x^{4-1})+12(2x^{2-1})\\ \frac{dy}{dx}&=42x^5-36x^3+24x\\ \end{align*}

Example 6:

$$ y=3x^3-4x^2+7x-10,\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y&=3x^3-4x^2+7x-10\\ \frac{dy}{dx}&=\frac{d}{dx}(3x^3-4x^2+7x-10)\\ \frac{dy}{dx}&=\frac{d}{dx}3x^3-\frac{d}{dx}4x^2+\frac{d}{dx}7x-\frac{d}{dx}10\\ \frac{dy}{dx}&=3\frac{d}{dx}x^3-4\frac{d}{dx}x^2+7\frac{d}{dx}x-\frac{d}{dx}10\\\\ \because\;\; \frac{d}{dx}x^n&=nx^{n-1}\,,\,\frac{d}{dx}x=1\,\textup{and}\,\frac{d}{dx}c=0\\\\ \therefore\;\; \frac{dy}{dx}&=3(3x^{3-1})-4(2x^{2-1})+7(1)-0\\ \frac{dy}{dx}&=9x^2-8x+7\\ \end{align*}

Example 7:

$$ y=\frac{1}{x},\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y&=\frac{1}{x}\\ \frac{dy}{dx}&=\frac{d}{dx}\left(\frac{1}{x}\right)\\ \frac{dy}{dx}&=\frac{d}{dx}x^{-1}\\\\ \because \frac{d}{dx}x^n&=nx^{n-1}\\\\ \therefore \frac{dy}{dx}&=-1x^{-1-1}\\ \frac{dy}{dx}&=-1x^{-2}\\ \frac{dy}{dx}&=\frac{-1}{x^2}\\ \frac{dy}{dx}&=-\frac{1}{x^2}\\ \end{align*}

Example 8:

$$ y=\frac{1}{x^2},\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y&=\frac{1}{x^2}\\ \frac{dy}{dx}&=\frac{d}{dx}\left(\frac{1}{x^2}\right)\\ \frac{dy}{dx}&=\frac{d}{dx}x^{-2}\\\\ \because \frac{d}{dx}x^n&=nx^{n-1}\\\\ \therefore \frac{dy}{dx}&=-2x^{-2-1}\\ \frac{dy}{dx}&=-2x^{-3}\\ \frac{dy}{dx}&=\frac{-2}{x^3}\\ \frac{dy}{dx}&=-\frac{2}{x^3}\\ \end{align*}

Example 9:

$$ y=\frac{4}{x^3},\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y&=\frac{4}{x^3}\\ \frac{dy}{dx}&=\frac{d}{dx}\left(\frac{4}{x^3}\right)\\ \frac{dy}{dx}&=\frac{d}{dx}4x^{-3}\\ \frac{dy}{dx}&=4\frac{d}{dx}x^{-3}\\\\ \because \frac{d}{dx}x^n&=nx^{n-1}\\\\ \therefore \frac{dy}{dx}&=4(-3x^{-3-1})\\ \frac{dy}{dx}&=-12x^{-4}\\ \frac{dy}{dx}&=\frac{-12}{x^4}\\ \frac{dy}{dx}&=-\frac{12}{x^4}\\ \end{align*}

Example 10:

$$ y=12x+\frac{3}{x},\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y&=12x+\frac{3}{x}\\ \frac{dy}{dx}&=\frac{d}{dx}\left(12x+\frac{3}{x}\right)\\ \frac{dy}{dx}&=\frac{d}{dx}12x+\frac{d}{dx}\frac{3}{x}\\ \frac{dy}{dx}&=\frac{d}{dx}12x+\frac{d}{dx}3x^{-1}\\ \frac{dy}{dx}&=12\frac{d}{dx}x+3\frac{d}{dx}x^{-1}\\\\ \because\;\; \frac{d}{dx}x&=1\;\; \textup{and} \;\:\frac{d}{dx}x^n=nx^{n-1}\\\\ \therefore\;\; \frac{dy}{dx}&=12(1)+3(-1x^{-1-1})\\ \frac{dy}{dx}&=12+3(-1x^{-2})\\ \frac{dy}{dx}&=12+3\left(\frac{-1}{x^2}\right)\\ \frac{dy}{dx}&=12-\frac{3}{x^2}\\ \end{align*}