How do you implicitly differentiate  sin x - cos y = e^(xy)sin x cos y?

Dec 16, 2017

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

Explanation:

use trig differentiation rules, e^x differentiation, chain rule, and product rule

$\frac{d}{\mathrm{dx}} \left(\sin x\right) - \frac{d}{\mathrm{dx}} \left(\cos y\right) = \frac{d}{\mathrm{dx}} \left({e}^{x y} \sin x \cos y\right)$

$\cos x + \sin y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = \left(\sin x \cos y\right) \left(\frac{d}{\mathrm{dx}} \left({e}^{x y}\right)\right) + \left({e}^{x y}\right) \frac{d}{\mathrm{dx}} \left(\sin x \cos y\right)$

$\cos x + \sin y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = \left(\sin x \cos y\right) \left({e}^{x y} \cdot \frac{d}{\mathrm{dx}} \left(x y\right)\right) + \left({e}^{x y}\right) \left(\cos y \cdot \frac{d}{\mathrm{dx}} \left(\sin x\right) + \sin x \cdot \frac{d}{\mathrm{dx}} \left(\cos y\right)\right)$

cosx+siny(dy/dx)=(sinxcosy)(e^(xy)*(y+x(dy/dx))+(e^(xy))(cosy*cosx+sinx*(-siny)*dy/dx)

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

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

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

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

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