# What is the implicit derivative of 25=sin(xy)/x-3xy?

Nov 19, 2015

Long time since tried this but I am having a go!
$\textcolor{b l u e}{\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sin \left(x y\right) - y x \cos \left(x y\right)}{{x}^{2} \cos \left(x y\right) - 3 {x}^{3}} + \frac{3 y}{\cos \left(x y\right) - 3 x}}$
$\textcolor{red}{\text{You will need to check this!!}}$

#### Explanation:

$\frac{d}{\mathrm{dx}} \left(\frac{\sin \left(x y\right)}{x} - 3 x y\right) = \frac{d}{\mathrm{dx}} \left(25\right)$

$\frac{d}{\mathrm{dx}} \left(\frac{\sin \left(x y\right)}{x}\right) - \frac{d}{\mathrm{dx}} \left(3 x y\right) = 0$

Using std forms $\frac{d}{\mathrm{dx}} \left(u v\right) = v \frac{\mathrm{du}}{\mathrm{dx}} + u \frac{\mathrm{dv}}{\mathrm{dx}}$

$\textcolor{w h i t e}{\times \times \times \times \times} \frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Let $\textcolor{w h i t e}{\times} w = u v \to x y$
then ${\textcolor{g r e e n}{\frac{\mathrm{dw}}{\mathrm{dx}} \to \frac{\mathrm{dy}}{\mathrm{dx}} = y + x \frac{\mathrm{dy}}{\mathrm{dx}} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left(1\right)}}_{\text{confirmed}}$

Let $t = \sin \left(x y\right) = \sin \left(w\right)$
then ${\textcolor{g r e e n}{\frac{\mathrm{dt}}{\mathrm{dx}} \to \frac{\mathrm{dy}}{\mathrm{dx}} = \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right) \cos \left(x y\right) \ldots \ldots \ldots \ldots . \left(2\right)}}_{\text{confirmed}}$

color(green)(d/dx(3xy) = 3(y+x(dy)/dx) "from (1) "................(3))_"confirmed"
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider: $\frac{d}{\mathrm{dx}} \left(\frac{\sin \left(x y\right)}{x}\right) \to \frac{u}{v}$

$\frac{x \left\{y \cos \left(x y\right) + x \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) \cos \left(x y\right)\right\} - \sin \left(x y\right)}{x} ^ 2 \ldots \ldots {.}_{\textcolor{red}{\text{corrected y-> ycos(xy)}}}$

${\textcolor{g r e e n}{\frac{y \cos \left(x y\right)}{x} + \frac{\mathrm{dy}}{\mathrm{dx}} \cos \left(x y\right) - \sin \frac{x y}{x} ^ 2. \ldots \ldots . . \left(4\right)}}_{\textcolor{red}{\left(\text{corrected} y \to y \cos \left(x y\right)\right)}}$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Combining (3) and (4)}}$

$\left\{\frac{y}{x} \cos \left(x y\right) + \frac{\mathrm{dy}}{\mathrm{dx}} \cos \left(x y\right) - \sin \frac{x y}{x} ^ 2\right\} - \left\{3 y + 3 x \frac{\mathrm{dy}}{\mathrm{dx}}\right\} = {0}_{\text{corrected } y \to y \cos \left(x y\right)}$

Collecting like terms $\textcolor{red}{\text{(rebuilt calculations)}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} \cos \left(x y\right) - 3 x \frac{\mathrm{dy}}{\mathrm{dx}} \textcolor{red}{=} \frac{\sin \left(x y\right)}{x} ^ 2 - \frac{y}{x} \cos \left(x y\right) + 3 {y}_{\textcolor{red}{\text{ corrected } y \to y \cos \left(x y\right)}}$
$\textcolor{w h i t e}{x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} \textcolor{red}{=} \frac{1}{\cos \left(x y\right) - 3 x} \left(\frac{\sin \left(x y\right)}{x} ^ 2 - \frac{y}{x} \cos \left(x y\right) + 3 y\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} \textcolor{red}{=} \frac{1}{\cos \left(x y\right) - 3 x} \left(\frac{\sin \left(x y\right) - y x \cos \left(x y\right)}{x} ^ 2 + 3 y\right)$

$\textcolor{b l u e}{\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sin \left(x y\right) - y x \cos \left(x y\right)}{{x}^{2} \cos \left(x y\right) - 3 {x}^{3}} + \frac{3 y}{\cos \left(x y\right) - 3 x}}$

Nov 19, 2015

By working step-wise taking derivatives of individual components you should end up with
(dy)/(dx) = (xycos(xy)-sin(xy)-3x^2y)/(3x^3-x^2cos(xy)

#### Explanation:

This is complex so check carefully before assuming what follows is valid:

Breaking the equation up into small pieces and working on the derivatives for each:
Part 1: The $\textcolor{red}{x y}$ component of $25 = \sin \frac{\textcolor{red}{x y}}{x} - 3 x y$
Finding the derivative of $x y$ is of the form for finding the derivative of $u v$ and we know $\frac{d \left(u v\right)}{\mathrm{dx}} = v \frac{\mathrm{du}}{\mathrm{dx}} + u \frac{\mathrm{dv}}{\mathrm{dx}}$
so
$\textcolor{w h i t e}{\text{XXX}} \frac{d \left(x y\right)}{\mathrm{dx}} = y \frac{\cancel{\mathrm{dx}}}{\cancel{\mathrm{dx}}} + x \frac{\mathrm{dy}}{\mathrm{dx}}$
$\textcolor{w h i t e}{\text{XXXXXX}} = y + x \frac{\mathrm{dy}}{\mathrm{dx}}$

Part 2: The $\textcolor{red}{\sin \left(x y\right)}$ component of $25 = \frac{\textcolor{red}{\sin \left(x y\right)}}{x} - 3 x y$
Finding the derivative of $\sin \left(x y\right)$ is of the form for finding the derivative of $f \left(g \left(u\right)\right)$ and we know
$\textcolor{w h i t e}{\text{XXX}} \frac{\mathrm{df} \left(g \left(x\right)\right)}{\mathrm{dx}} = \frac{\mathrm{df} \left(g \left(x\right)\right)}{d \left(g \left(x\right)\right)} \cdot \frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}}$
with $f \left(x\right) = \sin \left(x\right)$ and $g \left(x\right) = x y$
so
$\textcolor{w h i t e}{\text{XXX}} \frac{\mathrm{ds} \in \left(x y\right)}{\mathrm{dx}} = \cos \left(x y\right) \cdot \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

Part 3: The $\textcolor{red}{\sin \frac{x y}{x}}$ component of $25 = \textcolor{red}{\sin \frac{x y}{x}} - 3 x y$
Finding the derivative of $\sin \frac{x y}{x}$ is of the form for finding the derivative of $\frac{u}{v}$ and we know
$\textcolor{w h i t e}{\text{XXX}} d \frac{\frac{u}{v}}{\mathrm{dx}} = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{{v}^{2}}$
with $v = x$ and $u = \sin \left(x y\right)$
so
$\textcolor{w h i t e}{\text{XXX}} \frac{d \frac{\sin \left(x y\right)}{x}}{\mathrm{dx}} = \frac{x \cdot \left(\cos \left(x y\right) \cdot \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right)\right) - \sin \left(x y\right) \frac{\cancel{\mathrm{dx}}}{\cancel{\mathrm{dx}}}}{{x}^{2}}$

$\textcolor{w h i t e}{\text{XXXXXXXX}} = \frac{\cos \left(x y\right) \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right)}{x} - \frac{\sin \left(x y\right)}{{x}^{2}}$

Part 4: The $\textcolor{red}{x y}$ component of $25 = \sin \frac{x y}{x} - 3 \textcolor{red}{x y}$
from Part 1 we already have
$\textcolor{w h i t e}{\text{XXX}}$(d(xy))/(dx) = y+x(dy)/(dx)

Putting the pieces together
Take the derivative of both sides
$d \frac{25}{\mathrm{dx}} = \frac{d \left(\sin \frac{x}{x} - 3 x y\right)}{\mathrm{dx}}$

$0 = \frac{\mathrm{ds} \in \frac{x}{x}}{\mathrm{dx}} - 3 \left(\frac{d \left(x y\right)}{\mathrm{dx}}\right)$

$\frac{\cos \left(x y\right) \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right)}{x} - \frac{\sin \left(x y\right)}{{x}^{2}} - 3 \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

$\frac{y \cos \left(x y\right)}{x} + \cos \left(x y\right) \frac{\mathrm{dy}}{\mathrm{dx}} - \sin \frac{x y}{{x}^{2}} - 3 y - 3 x \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\frac{y \cos \left(x y\right)}{x} - \sin \frac{x y}{{x}^{2}} - 3 y = \left(3 x - \cos \left(x y\right)\right) \frac{\mathrm{dy}}{\mathrm{dx}}$

(dy)/(dx) = (xycos(xy)-sin(xy)-3x^2y)/(3x^3-x^2cos(xy)#