Chain Rule of Differentiation

Example 1:

$$ y=(x+5)^6,\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y&=(x+5)^6\\ \frac{dy}{dx} &=\frac{d}{dx} (x+5)^6\\\\ \textup{Let}\;\;\; u &= x+5 \\\\ \frac{dy}{dx} &=\frac{d}{dx} u^6\\ \frac{dy}{dx} &=\frac{d}{du} u^6 \frac{du}{dx}\\\\ \because\;\; \frac{d}{dx} x^n &=nx^{n-1}\\\\ \therefore \frac{dy}{dx} &=6u^{6-1}\frac{d}{dx}(x+5)\\ \frac{dy}{dx} &=6u^5(\frac{d}{dx}x+\frac{d}{dx}5)\\\\ \because\frac{d}{dx} x &=1 \;\;\; \textup{and} \;\;\;\frac{d}{dx}c=0\\\\ \therefore\frac{dy}{dx} &=6u^5(1+0)\\ \frac{dy}{dx} &=6u^5(1)\\ \frac{dy}{dx} &=6u^5\\ \frac{dy}{dx} &=6(x+5)^5\\ \end{align*}

Example 2:

$$ y=(3x-4)^5,\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y&=(3x-4)^5\\ \frac{dy}{dx}&=\frac{d}{dx} (3x-4)^5\\\\ \textup{Let}\;\;\; u &= 3x-4\\\\ \frac{dy}{dx}&=\frac{d}{dx} u^5\\ \frac{dy}{dx}&=\frac{d}{du} u^5 \frac{du}{dx}\\\\ \because\;\; \frac{d}{dx} x^n&=nx^{n-1}\\\\ \therefore \frac{dy}{dx}&=5u^{5-1}\frac{d}{dx}(3x-4)\\ \frac{dy}{dx}&=5u^4\left(\frac{d}{dx}3x-\frac{d}{dx}4\right)\\ \frac{dy}{dx}&=5u^4\left(3\frac{d}{dx}x-\frac{d}{dx}4\right)\\\\ \because\frac{d}{dx} x&=1 \;\;\; \textup{and} \;\;\;\frac{d}{dx}c=0\\\\ \therefore\frac{dy}{dx}&=5u^4(3(1)-0)\\ \frac{dy}{dx}&=5u^4(3)\\ \frac{dy}{dx}&=15u^4\\ \frac{dy}{dx}&=15(3x-4)^4\\ \end{align*}

Example 3:

$$ y=(2-5x)^4,\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y&=(2-5x)^4\\ \frac{dy}{dx}&=\frac{d}{dx} (2-5x)^4\\\\ \textup{Let}\;\;\; u &= 2-5x \\\\ \frac{dy}{dx}&=\frac{d}{dx} u^4\\ \frac{dy}{dx}&=\frac{d}{du} u^4 \frac{du}{dx}\\\\ \because\;\; \frac{d}{dx} x^n&=nx^{n-1}\\\\ \therefore \frac{dy}{dx}&=4u^{4-1}\frac{d}{dx}(2-5x)\\ \frac{dy}{dx}&=4u^3\left(\frac{d}{dx}2-\frac{d}{dx}5x\right)\\ \frac{dy}{dx}&=4u^3\left(\frac{d}{dx}2-5\frac{d}{dx}x\right)\\\\ \because\frac{d}{dx}c&=0 \;\;\; \textup{and} \;\;\;\frac{d}{dx} x=1\\\\ \therefore\frac{dy}{dx}&=4u^3(0-5(1))\\ \frac{dy}{dx}&=4u^3(-5)\\ \frac{dy}{dx}&=-20u^3\\ \frac{dy}{dx}&=-20(2-5x)^3\\ \end{align*}

Example 4:

$$ y=(4x^2+1)^7,\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y&=(4x^2+1)^7\\ \frac{dy}{dx}&=\frac{d}{dx}(4x^2+1)^7\\\\ \textup{Let}\;\;\; u &= 4x^2+1\\\\ \frac{dy}{dx}&=\frac{d}{dx} u^7\\ \frac{dy}{dx}&=\frac{d}{du} u^7 \frac{du}{dx}\\\\ \because\;\; \frac{d}{dx} x^n&=nx^{n-1}\\\\ \therefore \frac{dy}{dx}&=7u^{7-1}\frac{d}{dx}(4x^2+1)\\ \frac{dy}{dx}&=7u^6\left(\frac{d}{dx}4x^2+\frac{d}{dx}1\right)\\ \frac{dy}{dx}&=7u^6\left(4\frac{d}{dx}x^2+\frac{d}{dx}1\right)\\\\ \because\frac{d}{dx} x^n&=nx^{n-1} \;\;\; \textup{and} \;\;\;\frac{d}{dx}c=0\\\\ \therefore\frac{dy}{dx}&=7u^6(4(2x^{2-1})+0)\\ \frac{dy}{dx}&=7u^6(4(2x))\\ \frac{dy}{dx}&=56xu^6\\ \frac{dy}{dx}&=56x(4x^2+1)^6\\ \end{align*}

Example 5:

$$ y=(x^2+3x+2)^6,\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y&=(x^2+3x+2)^6\\ \frac{dy}{dx}&=\frac{d}{dx}(x^2+3x+2)^6\\\\ \textup{Let}\;\;\; u &= x^2+3x+2\\\\ \frac{dy}{dx}&=\frac{d}{dx} u^6\\ \frac{dy}{dx}&=\frac{d}{du} u^6 \frac{du}{dx}\\\\ \because\;\; \frac{d}{dx} x^n&=nx^{n-1}\\\\ \therefore \frac{dy}{dx}&=6u^{6-1}\frac{d}{dx}(x^2+3x+2)\\ \frac{dy}{dx}&=6u^5\left(\frac{d}{dx}x^2+\frac{d}{dx}3x+\frac{d}{dx}2\right)\\ \frac{dy}{dx}&=6u^5\left(\frac{d}{dx}x^2+3\frac{d}{dx}x+\frac{d}{dx}2\right)\\\\ \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}&=6u^5(2x^{2-1}+3(1)+0)\\ \frac{dy}{dx}&=6u^5(2x+3)\\ \frac{dy}{dx}&=6(2x+3)u^5\\ \frac{dy}{dx}&=6(2x+3)(x^2+3x+2)^5\\ \end{align*}

Example 6:

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

Solution:

\begin{align*} y&=\frac{1}{x+1}\\ \frac{dy}{dx}&=\frac{d}{dx} \frac{1}{x+1}\\\\ \textup{Let}\;\;\; u &= x+1\\\\ \frac{dy}{dx}&=\frac{d}{dx} \frac{1}{u}\\ \frac{dy}{dx}&=\frac{d}{dx} u^{-1}\\ \frac{dy}{dx}&=\frac{d}{du} u^{-1} \frac{du}{dx}\\\\ \because\;\; \frac{d}{dx} x^n&=nx^{n-1}\\\\ \therefore \frac{dy}{dx}&=-1u^{-1-1}\frac{d}{dx}(x+1)\\ \frac{dy}{dx}&=-1u^{-2}\left(\frac{d}{dx}x+\frac{d}{dx}1\right)\\\\ \because\frac{d}{dx} x&=1 \;\;\; \textup{and} \;\;\;\frac{d}{dx}c=0\\\\ \therefore\frac{dy}{dx}&=\frac{-1}{u^2}(1+0)\\ \frac{dy}{dx}&=-\frac{1}{u^2}(1)\\ \frac{dy}{dx}&=-\frac{1}{u^2}\\ \frac{dy}{dx}&=-\frac{1}{(x+1)^2}\\ \end{align*}

Example 7:

$$ y=\frac{1}{2x+3},\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

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