Implicit Differentiation

Section 3.4 Implicit Differentiation

When \(y\) is not solved explicitly it is called implicit function.

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For example: \(x^2 + y^2 = 25\) (equation of a circle).

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Differentiate both sides with respect to \(x\text{:}\)

\begin{equation*} 2x + 2y \frac{\,dy}{\,dx} = 0 \end{equation*}

\begin{equation*} \therefore \, \frac{\,dy}{\,dx} =y' = -\frac{x}{y} \end{equation*}

At point \((3, 4)\) on the circle: slope = \(-3/4.\)

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Example 3.4.1.

\begin{equation*} ax^2+2hxy+by^2+2gx+2fy+c=0 \end{equation*}

find \(\frac{dy}{dx.}\)

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Solution.

differentiate w. r. t. "x", we have -

\begin{align*} 2ax+2h(x\frac{dy}{dx}+y)+2by\frac{dy}{dx}+2g+2f\frac{dy}{dx} \amp =0 \\ \frac{dy}{dx} \left(2hx+2by+2f\right) \amp =-2ax-2hy-2g\\ or, \, \frac{dy}{dx} \amp = \frac{-2(ax+hy+g)}{2(hx+by+f)}\\ \therefore \, \frac{dy}{dx} \amp = \frac{-(ax+hy+g)}{(hx+by+f)} \end{align*}

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Example 3.4.2.

\begin{equation*} \frac{d}{dx}\sin y = \cos y \frac{dy}{dx} \end{equation*}

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\begin{equation*} \frac{d}{dx}\cos y^2 = -\sin y^2 2y \frac{dy}{dx} \end{equation*}

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\begin{equation*} \frac{d}{dx}y^3 = 3y^2 \frac{dy}{dx}. \end{equation*}

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Example 3.4.3.

\begin{equation*} \sqrt{x}=\sqrt{y} =4, \end{equation*}

find \(\frac{dx}{dy} \) at \(y=1.\)

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Solution.

\begin{equation*} \frac{1}{2\sqrt{x}}+ \frac{1}{2\sqrt{y}}\frac{dy}{dx} =0 \end{equation*}

\begin{equation*} \frac{dy}{dx} = -\frac{1}{2\sqrt{x}}\times 2\sqrt{y} = -\sqrt{\frac{y}{x}}. \end{equation*}

\begin{equation*} \therefore \, \frac{dx}{dy}\bigg|_{y=1} = \frac{1}{\frac{dx}{dy}\bigg|_{y=1}} = -\sqrt{\frac{x}{y}}\bigg|_{y=1} = -\sqrt{x}. \end{equation*}

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Example 3.4.4.

Find \(y'\) if \(x^3+y^3 =6xy.\)
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Find the tangent at the point \((3,3).\)
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At what point in first quadrant is the tangent line horizontal.
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Solution.

\begin{align*} x^3+y^3 \amp =6xy \\ 3x^2+3y^2\frac{dy}{dx} \amp =6x\frac{dy}{dx}+6y \end{align*}

on differentiating w.r.t. "x"

\begin{align*} x^3+y^3 \amp =6xy \\ (3y^2-6y) \frac{dy}{dx} \amp =(6y-3x^2)\\ or, \, \frac{dy}{dx}\amp = \frac{(6y-3x^2)}{(3y^2-6x)} = \frac{(2y-x^2)}{(y^2-2x)} \end{align*}

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at \(x=3,\) and \(y=3\)

\begin{align*} y' = \frac{dy}{dx} \amp = \frac{(2\times 3-3^2)}{(3^2-2\times 3)} = -1. \end{align*}

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Now, the equation of tangent line is

\begin{align*} y-y_1 \amp = m(x-x_1) \\ y-3 \amp =-1(x-3)\\ y \amp = -x+3+3 = -x+6\\ x+y \amp =6. \end{align*}

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The tangent line is horzontal if slope, \(y'=0\text{,}\)

\begin{align*} y' \amp =0 \\ \frac{(2y-x^2)}{(y^2-2x)} \amp =0 \\ 2y-x^2 \amp =0 \, [\text{ provided } \, y^2-2x \neq 0 ]\\ \therefore \, y \amp =\frac{x^2}{2}. \end{align*}

From equation \(x^3+y^3 =6xy\text{,}\) we have

\begin{align*} x^3+(\frac{x^2}{2})^3 \amp =6x(\frac{x^2}{2}) \\ x^3+\frac{x^6}{8} \amp =3x^3\\ x^3 \amp = \frac{x^6}{16}\\ x^3 \amp =16\\ x \amp =16^{1/3} = 2^{\frac{4}{3}} \end{align*}

\begin{equation*} Hence, \, y=\frac{1}{2}(2^{4/3})^2 = 2^{5/3} \end{equation*}

Therefore the tangent is horizontal at \((2^{4/3}, 2^{5/3})\) in first quadrant.

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Example 3.4.5.

\begin{equation*} \tan \left(\frac{x}{y}\right) = x+y \end{equation*}

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Solution.

Differentiate w.r.t. "x",

\begin{align*} \sec^2 \left(\frac{x}{y}\right) \left[\frac{1}{y}+x(-y^{-2}\frac{dy}{dx})\right] \amp=1+\frac{dy}{dx} \\ \left[\frac{1}{y}-\frac{x}{y^2}\frac{dy}{dx}\right] \sec^2 \left(\frac{x}{y}\right) \amp = 1+\frac{dy}{dx} \\ -\frac{dy}{dx} \left[1+ \frac{x}{y^2}\right]\sec^2 \left(\frac{x}{y}\right) \amp = 1- \frac{1}{y}\sec^2 \left(\frac{x}{y}\right)\\ \frac{dy}{dx} \amp = \frac{\left[\frac{1}{y}\sec^2 \left(\frac{x}{y}\right)-1 \right]}{1+\left[\frac{x}{y^2}\sec^2 \left(\frac{x}{y}\right)\right]}\\ y' \amp = \frac{\left[y\sec^2 \left(\frac{x}{y}\right)-y^2 \right]}{\left[ y^2+x\sec^2 \left(\frac{x}{y}\right)\right]} \end{align*}

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Subsection 3.4.1 Parametric Functions

parameteric equation for a circle:

\begin{equation*} x^2+y^2=r^2 \end{equation*}

where, \(x= r\cos\theta,\) and \(y=r\sin\theta\text{.}\)

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parameteric equation for a parabola:

\begin{equation*} y^2=4 a x \end{equation*}

where, \(x= at^2,\) and \(y=2at\text{.}\)

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If \(y=f(t)\) and \(x=g(t)\) then, find \(\frac{dy}{dx}\text{.}\) Here

\begin{equation*} \frac{dy}{dt}=f'(t) \, \text{ and } \, \frac{dx}{dt}=g'(t) \end{equation*}

\begin{equation*} \therefore \, \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{dy}{dt} \times \frac{dt}{dx} =\frac{f'(t)}{g'(t)}. \end{equation*}

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Example 3.4.6.

\begin{equation*} x=\frac{2t}{1+t^2}, \quad \, y= \frac{1-t^2}{1+t^2} \,\quad \text{ at } t=2. \end{equation*}

Find \(\frac{dy}{dx}.\)

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Solution.

\begin{equation*} \frac{dx}{dt} =\frac{(1+t^2)^2\cdot 2- 2t(2t)}{(1+t^2)^2} = \frac{2-2t^2}{(1+t^2)^2} = \frac{2(1-t^2)}{(1+t^2)^2} \end{equation*}

\begin{equation*} \frac{dy}{dt} =\frac{(1+t^2)(-2t)- (1-t^2)(2t)}{(1+t^2)^2} = \frac{-4t}{(1+t^2)^2} \end{equation*}

\begin{equation*} \therefore \, \frac{dy}{dx} = \frac{\frac{-4t}{(1+t^2)^2} }{ \frac{2(1-t^2)}{(1+t^2)^2}} \end{equation*}

\begin{equation*} = \frac{-4t}{(1+t^2)^2}\times \frac{(1+t^2)^2}{2(1-t^2) } = \frac{-2t}{(1-t^2)} \end{equation*}

\begin{equation*} and, \, \frac{dy}{dx}\bigg|_{t=2} = \frac{-2\times 2}{(1-2^2)} = \frac{4}{3}. \end{equation*}

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Example 3.4.7.

\begin{equation*} x=t^2, \, y = 2t, \end{equation*}

find \(\frac{dy}{dx}\bigg|_{(1,2)}\)

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Solution.

\begin{equation*} \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2}{2t} =\frac{1}{t} \end{equation*}

\begin{equation*} \therefore \, \frac{dy}{dx}\bigg|_{(1,2)} = \frac{1}{t} \end{equation*}

Since, at (1,2)

\begin{align*} x \amp =t^2\\ 1 \amp = t^2 \, \Rightarrow t=\pm 1. \end{align*}

and

\begin{align*} y \amp =2t\\ 2 \amp = 2t \, \Rightarrow t=\pm 1 \end{align*}

Hence, \(t=1.\)

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Example 3.4.8.

Find \(\frac{dy}{dx}\text{,}\) if \(x=at^2;\) and \(y=2at\text{.}\)

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Solution.

\begin{equation*} \frac{dx}{dt} = 2at \, \quad \, \frac{dy}{dt} = 2a \end{equation*}

\begin{equation*} \therefore \, \frac{dy}{dx} = \frac{2a}{2at} =\frac{1}{t}. \end{equation*}

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