Chain Rule Differentiation

Chain Rule Differentiation

In differential calculus, the chain rule is a formula for determining the derivative of the combined two or more functions. the chain rule could be used in Leibniz’s notation in the following way.
If $y=g(u)$ where $u=f(x)$ then $\displaystyle \dfrac{dy}{dx}=\dfrac{dy}{du} \times \dfrac{du}{dx}$.
Generally, the chain rule is described as following in its simplest understanding.
If $y=\big[f(x)\big]^n$ then $\displaystyle \dfrac{dy}{dx}=n\big[f(x)\big]^{n-1}\times f'(x)$.

Example 1

Find $\displaystyle \dfrac{dy}{dx}$ of $y=(x^2-4x)^5$.

\( \begin{align} \displaystyle
\dfrac{dy}{dx} &= 5(x^2-4x)^{5-1} \times \dfrac{d}{dx}(x^2-4x) \\
&= 5(x^2-4x)^4 (2x-4) \\
&= 10(x^2-4x)^4 (x-2)
\end{align} \)

Example 2

Find $\displaystyle \dfrac{dy}{dx}$ of $y=\dfrac{1}{(x^3-3)^6}$.

\( \begin{align} \displaystyle
\dfrac{1}{(x^3-3)^6} &= (x^3-3)^{-6} \\
\dfrac{dy}{dx} &= -6(x^3-3)^{-6-1} \times \dfrac{d}{dx}(x^3-3) \\
&= -6(x^3-3)^{-7} \times 3x^2 \\
&= \dfrac{-18x^2}{(x^3-3)^7}
\end{align} \)

Example 3

Find $\displaystyle \dfrac{dy}{dx}$ of $y=\sqrt{3x-1}$.

\( \begin{align} \displaystyle
\sqrt{3x-1} &= (3x-1)^{\frac{1}{2}} \\
\dfrac{dy}{dx} &= \dfrac{1}{2} (3x-1)^{\frac{1}{2}-1} \times \dfrac{d}{dx}(3x-1) \\
&= \dfrac{1}{2} (3x-1)^{-\frac{1}{2}} \times 3 \\
&= \dfrac{3}{2\sqrt{3x-1}}
\end{align} \)

Example 4

Find $\displaystyle \dfrac{dy}{dx}$ of $y=\sqrt[3]{x^2-x}$.

\( \begin{align} \displaystyle
\sqrt[3]{x^2-x} &= (x^2-x)^{\frac{1}{3}} \\
\dfrac{dy}{dx} &= \dfrac{1}{3}(x^2-x)^{\frac{1}{3}-1} \times \dfrac{d}{dx}(x^2-x) \\
&= \dfrac{1}{3}(x^2-x)^{-\frac{2}{3}} \times (2x-1) \\
&= \dfrac{2x-1}{3\sqrt[3]{(x^2-x)^2}}
\end{align} \)

Example 5

Find $\displaystyle \dfrac{dy}{dx}$ of $y=\dfrac{1}{\sqrt{x^3-2x}}$.

\( \begin{align} \displaystyle
\dfrac{1}{\sqrt{x^3-2x}} &= (x^3-2x)^{-\frac{1}{2}} \\
\dfrac{dy}{dx} &= -\dfrac{1}{2}(x^3-2x)^{-\frac{1}{2}-1} \times (x^3-2x) \\
&= -\dfrac{1}{2}(x^3-2x)^{-\frac{3}{2}} \times (3x^2-2) \\
&= -\dfrac{3x^2-2}{2\sqrt{(x^3-2x)^3}}
\end{align} \)

Extension Examples

These Extension Examples require to have some prerequisite skills including;
\( \begin{align} \displaystyle
\dfrac{d}{dx}\sin{x} &= \cos{x} \\
\dfrac{d}{dx}\cos{x} &= -\sin{x} \\
\dfrac{d}{dx}e^x &= e^x \\
\dfrac{d}{dx}\log_e{x} &= \dfrac{1}{x}
\end{align} \)

Example 6

Find $\displaystyle \dfrac{dy}{dx}$ of $y=\sin^3{x}$, known that $\dfrac{d}{dx}\sin{x} = \cos{x}$.

\( \begin{align} \displaystyle
y &= (\sin{x})^3 \\
\dfrac{dy}{dx} &= 3(\sin{x})^{3-1} \times \dfrac{d}{dx}\sin{x} \\
&= 3(\sin{x})^{2} \times \cos{x} \\
&= 3\sin^2{x}\cos{x}
\end{align} \)

Example 7

Find $\displaystyle \dfrac{dy}{dx}$ of $y=\cos^3{x}$, known that $\dfrac{d}{dx}\cos{x} = -\sin{x}$.

\( \begin{align} \displaystyle
y &= (\cos{x})^3 \\
\dfrac{dy}{dx} &= 3(\cos{x})^{3-1} \times \dfrac{d}{dx}\cos{x} \\
&= 3(\cos{x})^{2} \times (-\sin{x}) \\
&= -3\cos^2{x}\sin{x}
\end{align} \)

Example 8

Find $\displaystyle \dfrac{dy}{dx}$ of $y=e^{\sin{x}}$, known that $\dfrac{d}{dx}\sin{x} = \cos{x}$ and $\dfrac{d}{dx}e^x = e^x$.

\( \begin{align} \displaystyle
\dfrac{dy}{dx} &= e^{\sin{x}} \times \dfrac{d}{dx}\sin{x} \\
&= e^{\sin{x}}\cos{x}
\end{align} \)

Example 9

Find $\displaystyle \dfrac{dy}{dx}$ of $y=\sin{e^{2x}}$, known that $\dfrac{d}{dx}\sin{x} = \cos{x}$ and $\dfrac{d}{dx}e^x = e^x$.

\( \begin{align} \displaystyle
\dfrac{dy}{dx} &= e^{\sin{e^{2x}}} \times \dfrac{d}{dx}e^{2x} \\
&= e^{\sin{e^{2x}}} \times e^{2x} \times \dfrac{d}{dx}2x \\
&= e^{\sin{e^{2x}}} \times e^{2x} \times 2 \\
&= 2e^{\sin{e^{2x}}} e^{2x}
\end{align} \)

Example 10

Find $\displaystyle \dfrac{dy}{dx}$ of $y=\log_e{\sin{e^{x^2}}}$, known that $\dfrac{d}{dx}\sin{x} = \cos{x}$ and $\dfrac{d}{dx}\log_e{x} = \dfrac{1}{x}$.

\( \begin{align} \displaystyle
\dfrac{d}{dx} &= \dfrac{1}{\sin{e^{x^2}}} \times \dfrac{d}{dx}\sin{e^{x^2}} \\
&= \dfrac{1}{\sin{e^{x^2}}} \times \cos{e^{x^2}} \times \dfrac{d}{dx} e^{x^2} \\
&= \dfrac{1}{\sin{e^{x^2}}} \times \cos{e^{x^2}} \times e^{x^2} \times \dfrac{d}{dx}x^2 \\
&= \dfrac{1}{\sin{e^{x^2}}} \times \cos{e^{x^2}} \times e^{x^2} \times 2x \\
&= \dfrac{2xe^{x^2}\cos{e^{x^2}}}{\sin{e^{x^2}}}
\end{align} \)

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