Calculus: Derivatives II

Implicit Differentiation

Not all equations come to us in the tidy form of $y=f(x)$. Some equations come as a jumble of both $x$ and $y$ terms. Such equations are called implicit equations because the dependent variable, $y$, is defined implicitly by the equation, rather than explicitly in the form of $y=f(x)$.

Solving for $y$ directly before differentiating can often be very difficult, but the good news is that one need not solve for $y$ directly in order to take the derivative. To differentiate a term involving $y$, simply differentiate it as if it were the independent variable, then multiply it by $\dfrac{dy}{dx}$. In other words, just apply the chain rule!

Problems

1. Differentiate with respect to $x$: $x^2 + y^2 = r^2$
$\dfrac{d}{dx}y^2 + \dfrac{d}{dx}x^2 = \dfrac{d}{dx}r^2 \\ 2y\dfrac{dy}{dx} + 2x = 0 \\ \dfrac{dy}{dx} = -\dfrac{x}{y}$
2. Differentiate with respect to $x$: $yx = 5$
$\dfrac{d}{dx}yx = \dfrac{d}{dx}5 \\ \left(\dfrac{d}{dx}y\right)x + y\left(\dfrac{d}{dx}x\right) = 0 \\ \dfrac{dy}{dx}x + y = 0 \\ \dfrac{dy}{dx} = \dfrac{-y}{x} \\$
3. Differentiate with respect to $x$: $yx = 5y$

$\dfrac{d}{dx}yx = \dfrac{d}{dx}5y \\ \left(\dfrac{d}{dx}y\right)x + y\left(\dfrac{d}{dx}x\right) = 5\dfrac{dy}{dx} \\ \left(\dfrac{dy}{dx}\right)x + y\left(1\right) = 5\dfrac{dy}{dx} \\ \dfrac{dy}{dx}x + y = 5\dfrac{dy}{dx} \\ \dfrac{dy}{dx}x - 5\dfrac{dy}{dx} = -y \\ \dfrac{dy}{dx}(x - 5) = -y \\ \dfrac{dy}{dx} = \dfrac{-y}{x-5} \\$
4. Differentiate with respect to $x$: $y^6 + y^4 + y^2 = x - x^3 - x^5$
$\dfrac{d}{dx}\left(y^6 + y^4 + y^2\right) = \dfrac{d}{dx}\left(x - x^3 - x^5\right) \\ \dfrac{d}{dx}y^6 + \dfrac{d}{dx}y^4 + \dfrac{d}{dx}y^2 = \dfrac{d}{dx}x - \dfrac{d}{dx}x^3 - \dfrac{d}{dx}x^5 \\ 6y^5\dfrac{dy}{dx} + 4y^3\dfrac{dy}{dx} + 2y\dfrac{dy}{dx} = 1 - 3x^2 - 5x^4 \\ \dfrac{dy}{dx}\left(6y^5 + 4y^3 + 2y\right) = 1 - 3x^2 - 5x^4 \\ \dfrac{dy}{dx} = \dfrac{1-3x^2-5x^4}{6y^5 + 4y^3 + 2y} \\$
5. Differentiate with respect to $x$: $yx^2 + y^3 = 4x$
$\dfrac{d}{dx}yx^2 + \dfrac{d}{dx}y^3 = \dfrac{d}{dx}4x \\ \dfrac{dy}{dx}x^2 + 2xy + 3y^2\dfrac{dy}{dx} = 4 \\ \dfrac{dy}{dx}\left(x^2 + 3y^2\right) + 2xy = 4 \\ \dfrac{dy}{dx}\left(x^2 + 3y^2\right) = 4-2xy \\ \dfrac{dy}{dx} = \dfrac{4-2xy}{x^2+3y^2} \\$
6. Differentiate with respect to $x$: $\sin(2x) = \cos(3y)$

$\dfrac{d}{dx}\sin(2x) = \dfrac{d}{dx}\cos(3y) \\ 2\cos(2x) = -\sin(3y)\dfrac{d}{dx}(3y) \\ 2\cos(2x) = -3\sin(3y)\dfrac{dy}{dx} \\ \dfrac{dy}{dx} = -\dfrac{3\sin(3y)}{2\cos(2x)} \\$
7. Differentiate with respect to $x$: $e^{y} = \sin(x)$

$\dfrac{d}{dx}e^{y} = \dfrac{d}{dx}\sin(x) \\ e^y\dfrac{d}{dx}y = \cos(x) \\ e^y\dfrac{dy}{dx} = \cos(x) \\ \dfrac{dy}{dx} = \dfrac{\cos(x)}{e^y} \\$
8. Differentiate with respect to $x$: $2^{xy} = 4^{\cos(x)}$
Taking the derivative directly will be messy. We should instead simplify first:

$2^{xy} = 4^{\cos(x)} \\ 2^{xy} = 2^{2\cos(x)} \\ \log_{2}\left(2^{xy}\right) = \log_{2}\left(2^{2\cos(x)}\right) \\ xy = 2\cos(x) \\$

This looks easier:

$xy = 2\cos(x) \\ \dfrac{d}{dx}xy = \dfrac{d}{dx}2\cos(x) \\ \left(\dfrac{d}{dx}x\right)y + x\left(\dfrac{d}{dx}y\right) = -2\sin(x) \\ (1)y + x\left(\dfrac{dy}{dx}\right) = -2\sin(x) \\ y + x\dfrac{dy}{dx} = -2\sin(x) \\ x\dfrac{dy}{dx} = -2\sin(x) - y \\ \dfrac{dy}{dx} = \dfrac{-2\sin(x) - y}{x} \\$
9. Differentiate with respect to $\alpha$: $y\alpha + \ln(y\alpha) = 5\cos(\alpha)$
$\dfrac{d}{d\alpha}y\alpha + \dfrac{d}{d\alpha}\ln(y\alpha) = \dfrac{d}{d\alpha}5\cos(\alpha) \\ = \left(\dfrac{d}{d\alpha}y\right)\alpha + y\left(\dfrac{d}{d\alpha}\alpha\right) + \dfrac{1}{y\alpha}\left(\dfrac{d}{d\alpha}y\alpha\right) = -5\sin(\alpha) \\ = \left(\dfrac{dy}{d\alpha}\right)\alpha + y\left(1\right) + \dfrac{1}{y\alpha}\left(\left(\dfrac{d}{d\alpha}y\right)\alpha + y\left(\dfrac{d}{d\alpha}\alpha\right)\right) = -5\sin(\alpha) \\ = \dfrac{dy}{d\alpha}\alpha + y + \dfrac{1}{y\alpha}\left(\left(\dfrac{dy}{d\alpha}\right)\alpha + y\left(1\right)\right) = -5\sin(\alpha) \\ = \dfrac{dy}{d\alpha}\alpha + y + \dfrac{1}{y\alpha}\left(\dfrac{dy}{d\alpha}\alpha + y\right) = -5\sin(\alpha) \\ = \dfrac{dy}{d\alpha}\alpha + y + \dfrac{dy}{d\alpha}\dfrac{1}{y} + \dfrac{1}{\alpha} = -5\sin(\alpha) \\ = \dfrac{dy}{d\alpha}\left(\alpha +\dfrac{1}{y}\right)+ y + \dfrac{1}{\alpha} = -5\sin(\alpha) \\ = \dfrac{dy}{d\alpha}\left(\alpha +\dfrac{1}{y}\right) = -5\sin(\alpha) - y - \dfrac{1}{\alpha} \\ = \dfrac{dy}{d\alpha} = \dfrac{-5\sin(\alpha) - y - \dfrac{1}{\alpha}}{\alpha +\dfrac{1}{y}} \\ = \dfrac{dy}{d\alpha} = \dfrac{-5y\alpha\sin(\alpha) - y^2\alpha - y}{\alpha^2y +\alpha} \\$
10. Differentiate with respect to $x$: $\ln(y+x) = \tan(\sin(xy))$

$\dfrac{d}{dx}\ln(y+x) = \dfrac{d}{dx}\tan(\sin(xy)) \\ \dfrac{1}{y+x}\left(\dfrac{d}{dx}\left(y+x\right)\right) = \sec^2(\sin(xy))\dfrac{d}{dx}\sin(xy) \\ \dfrac{1}{y+x}\left(\dfrac{dy}{dx}+1\right) = \sec^2(\sin(xy))\cos(xy)\dfrac{d}{dx}(xy) \\ \dfrac{dy}{dx}\dfrac{1}{y+x} + \dfrac{1}{y+x} = \sec^2(\sin(xy))\cos(xy)\left(\left(\dfrac{d}{dx}x\right)y + x\left(\dfrac{d}{dx}y\right)\right) \\ \dfrac{dy}{dx}\dfrac{1}{y+x} + \dfrac{1}{y+x} = \sec^2(\sin(xy))\cos(xy)\left(\left(1\right)y + x\left(\dfrac{dy}{dx}\right)\right) \\ \dfrac{dy}{dx}\dfrac{1}{y+x} + \dfrac{1}{y+x} = \sec^2(\sin(xy))\cos(xy)\left(y + \dfrac{dy}{dx}x\right) \\ \dfrac{dy}{dx}\dfrac{1}{y+x} + \dfrac{1}{y+x} = y\sec^2(\sin(xy))\cos(xy) + \dfrac{dy}{dx}x\sec^2(\sin(xy))\cos(xy) \\ \dfrac{dy}{dx}\dfrac{1}{y+x} - \dfrac{dy}{dx}x\sec^2(\sin(xy))\cos(xy) = y\sec^2(\sin(xy))\cos(xy) - \dfrac{1}{y+x} \\ \dfrac{dy}{dx}\left(\dfrac{1}{y+x} - x\sec^2(\sin(xy))\cos(xy)\right) = y\sec^2(\sin(xy))\cos(xy) - \dfrac{1}{y+x} \\ \dfrac{dy}{dx} = \dfrac{y\sec^2(\sin(xy))\cos(xy) - \dfrac{1}{y+x}}{\dfrac{1}{y+x} - x\sec^2(\sin(xy))\cos(xy)} \\ \dfrac{dy}{dx} = \dfrac{y(x+y)\sec^2(\sin(xy))\cos(xy) - 1}{1 - x(x+y)\sec^2(\sin(xy))\cos(xy)} \\ \dfrac{dy}{dx} = -\dfrac{y(x+y)\sec^2(\sin(xy))\cos(xy) - 1}{x(x+y)\sec^2(\sin(xy))\cos(xy)-1} \\$