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Implicit Differentiation

Implicit Differentiation

The key idea behind implicit differentiation is that, when we are given an equation, $f(x,y) = k$, we can think of $y$ as some function of $x$. Once we think of y as a function of x, we can then differentiate functions of $y$ with respect to $x$ by using the chain rule. That is given a function $g$, expressed in terms of $y$, we can determine the derivative $\frac{dg}{dx}$ $$ \frac{d}{dx}\left(g(y)\right) = \frac{dg}{dx} = \frac{dg}{dy}\times \frac{dy}{dx} $$ This is easier to see through examples. For example, given $g(y) = y^2$, determine $\frac{dg}{dx}$. $$ \frac{dg}{dx} = \frac{dg}{dy}\times \frac{dy}{dx} = 2y\frac{dy}{dx} $$ Consider a more complex example. Let $h(y) = \sin(y^3)$. Then $$ \frac{dh}{dx} = \frac{dh}{dy}\times \frac{dy}{dx} =3y^2\cos(y^3)\frac{dy}{dx} $$ We also need to consider the case where we have the product of functions of $x$ and $y$, that is, differentiate the expression $p(x)q(y)$ with respect to $x$ $$ \begin{array}{rcl} \frac{d}{dx}\left(p(x)q(y)\right) &=& \frac{dp}{dx}q(y) + p(x)\frac{d}{dx}\left(q(y)\right) \\ &=& \frac{dp}{dx}q(y) + p(x)\frac{dq}{dy}\frac{dy}{dx} \end{array} $$ Again, consider the following examples. Differentiate the expression $y^2x^3$ with respect to $x$ $$ \begin{array}{rcl} \frac{d}{dx}\left(y^2x^3\right) &=& \frac{d}{dx}\left(x^3\right)y^2 + x^3\frac{d}{dx}\left(y^2\right) \\ &=&3x^2y^2 + x^3\frac{d(y^2)}{dy}\frac{dy}{dx} \\ &=& 3x^2y^2 + 2x^3y\frac{dy}{dx} \end{array} $$

Implicit Differentiation 1

Determine an expression for $\frac{dy}{dx}$ for the curve given by $$ x^2 + xy + y^2 = 10 $$ Express your answer in the form $\frac{dy}{dx} = f(x,y)$
solution - press button to display
$$\frac{d}{dx}(x^2) = 2x,\;\frac{d}{dx}(xy) = x\frac{dy}{dx} + y,\;\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$$ Therefore, $$ \begin{array}{rcl} \frac{d}{dx}\left(x^2 + xy + y^2\right) &=& \frac{d}{dx}\left(10\right) \\ 2x + x\frac{dy}{dx} + y + 2y\frac{dy}{dx} &=& 0 \\ \frac{dy}{dx} &=& -\frac{2x+y}{2y+x} \end{array} $$

Implicit Differentiation 2

Determine the coordinates of the stationary points of the curve with equation $$ 2x^2 - 2xy + y^2 = 16 $$
solution - press button to display
$$ \begin{array}{rcl} \frac{d}{dx}\left(2x^2 - 2xy + y^2 \right) &=& \frac{d}{dx}\left(16\right) \\ 4x -2x\frac{dy}{dx} - 2y + 2y\frac{dy}{dx} &=& 0 \\ \frac{dy}{dx} &=& \frac{2y-4x}{2y-2x} \end{array} $$ At the stationary points, $\frac{dy}{dx} = 0$. That is, $2y - 4x = 0 \Rightarrow y = 2x$ Substitution of this into the original equation yields $2x^2 - 2x(2x) + (2x)^2 = 16$. This gives $x^2 = 8 \rightarrow x = \pm2 \sqrt{2} \Rightarrow y = \pm 4 \sqrt{2}$ The image shows the curve, together with the two stationary points highlighted.