Chain Rule of Differentiation

Example 1:

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

Solution:

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

Example 2:

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

Solution:

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

Example 3:

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

Solution:

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

Example 4:

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

Solution:

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

Example 5:

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

Solution:

$$ y=(x^2+3x+2)^6$$ $$ \frac{dy}{dx}=\frac{d}{dx}(x^2+3x+2)^6$$ $$ \textup{Let}\;\;\; u = x^2+3x+2$$ $$ \frac{dy}{dx}=\frac{d}{dx} u^6$$ $$ \frac{dy}{dx}=\frac{d}{du} u^6 \frac{du}{dx}$$ $$ \because\;\; \frac{d}{dx} x^n=nx^{n-1}$$ $$ \therefore \frac{dy}{dx}=6u^{6-1}\frac{d}{dx}(x^2+3x+2)$$ $$ \frac{dy}{dx}=6u^5\left(\frac{d}{dx}x^2+\frac{d}{dx}3x+\frac{d}{dx}2\right)$$ $$ \frac{dy}{dx}=6u^5\left(\frac{d}{dx}x^2+3\frac{d}{dx}x+\frac{d}{dx}2\right)$$ $$ \because\frac{d}{dx} x^n=nx^{n-1} ,\;\;\frac{d}{dx}x=1\;\; \textup{and} \;\;\;\frac{d}{dx}c=0$$ $$ \therefore\frac{dy}{dx}=6u^5(2x^{2-1}+3(1)+0)$$ $$ \frac{dy}{dx}=6u^5(2x+3)$$ $$ \frac{dy}{dx}=6(2x+3)u^5$$ $$ \frac{dy}{dx}=6(2x+3)(x^2+3x+2)^5$$

Example 6:

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

Solution:

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

Example 7:

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

Solution:

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