This very powerful differentiation process follows from the chain rule.
$$u = g(f(x)) \
\frac{du}{dx} = g'(f(x)) \times f'(x)$$
We’ve done quite a few differentiation and derivatives, but all have been the differentiation of functions of the form \( y = f(x) \). Not all functions will fall into this simple form. The process that we are going to cover is called implicit differentiation.
The following examples require the use of implicit differentiation. The essential skill of implicit differentiation is a special case of the chain rule for derivatives. Let’s take a look at them now!
Question 1
Find \( \dfrac{dy}{dx} \) for \( xy = 1 \).
\( \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
\frac{d}{dx}xy &= \frac{d}{dx}1 \\
\frac{d}{dx}x \times y + x \times \frac{d}{dx}y &= 0 &\color{red} \text{product rule}\\
1 \times y + x\frac{dy}{dx} &= 0 \\
y + x\frac{dy}{dx} &= 0 \\
x\frac{dy}{dx} &= -y \\
\therefore \frac{dy}{dx} &= -\frac{y}{x}
\end{aligned} \)
Question 2
Find \( \dfrac{dy}{dx} \) for \( x^2 + y^2 = 1 \).
\( \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
\frac{d}{dx}x^2 + \frac{d}{dx}y^2 &= \frac{d}{dx}1 \\
2x + 2y \times \frac{dy}{dx} &= 0 &\color{red} \text{chain rule}\\
2y \times \frac{dy}{dx} &= -2x \\
\frac{dy}{dx} &= -\frac{2x}{2y} \\
\therefore \frac{dy}{dx} &= -\frac{x}{y}
\end{aligned} \)
Question 3
Find \( \dfrac{dy}{dx} \) for \( x^4y^5 = y + 1 \).
\( \begin{aligned} \displaystyle
\frac{d}{dx}x^4 \times y^5 + x^4 \times \frac{d}{dx} y^5 &= \frac{d}{dx}y + \frac{d}{dx}1 \\
4x^3y^5 + x^4 \times 5y\frac{dy}{dx} &= \frac{dy}{dx} + 0 \\
4x^3y^5 + 5x^4y\frac{dy}{dx} &= \frac{dy}{dx} \\
5x^4y\frac{dy}{dx} – \frac{dy}{dx} &= -4x^3y^5 \\
(5x^4y -1)\frac{dy}{dx} &= -4x^3y^5 \\
\therefore \frac{dy}{dx} &= -\frac{4x^3y^5}{5x^4y -1}
\end{aligned} \)
Question 4
Find \( \dfrac{dy}{dx} \) for \( \sin{y} = x \).
\( \begin{aligned} \displaystyle
\frac{d}{dx} \sin{y} &= \frac{d}{dx}x \\
\cos{y} \times \frac{dy}{dx} &= 1 \\
\therefore \frac{dy}{dx} &= \frac{1}{\cos{y}}
\end{aligned} \)
Question 5
Differentiate \( \cos{xy} \) in terms of \( x \).
\( \begin{aligned} \displaystyle
\frac{d}{dx} \cos{xy} &= – \sin{xy} \times \frac{d}{dx}xy \\
&= -\sin{xy} \times \Big(\frac{d}{dx}x \times y + x \times \frac{d}{dx}y\Big) \\
&= -\sin{xy} \times \Big(y + x\frac{dy}{dx}\Big)
\end{aligned} \)
Question 6
Find \( \dfrac{dy}{dx} \) for \( \cos^2{x} + \cos^2{y} = \cos{(x+y)} \).
\( \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
\frac{d}{dx}\cos^2{x} + \frac{d}{dx}\cos^2{y} &= \frac{d}{dx}\cos{(x+y)} \\
2 \cos{x} \times \frac{d}{dx} \cos{x} + 2 \cos{y} \times \frac{d}{dx}\cos{y} &= -\sin{(x+y)} \times \frac{d}{dx}(x+y)\\
2 \cos{x} \times (-\sin{x}) + 2 \cos{y} \times (- \sin{y}) \times \frac{dy}{dx} &= -\sin{(x+y)} \times \Big(1+\frac{dy}{dx}\Big) \\
-2 \sin{x} \cos{x}-2 \sin{y} \cos{y} \frac{dy}{dx} &= -\sin{(x+y)}-\sin{(x+y)} \frac{dy}{dx} \\
\sin{(x+y)} \frac{dy}{dx}-2 \sin{y} \cos{y} \frac{dy}{dx} &= 2 \sin{x} \cos{x}-\sin{(x+y)} \\
\left[\sin{(x+y)}-2 \sin{y} \cos{y}\right] \frac{dy}{dx} &= 2 \sin{x} \cos{x}-\sin{(x+y)} \\
\therefore \frac{dy}{dx} &= \frac{2 \sin{x} \cos{x}-\sin{(x+y)}}{\sin{(x+y)}-2 \sin{y} \cos{y}}
\end{aligned} \)
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