# How do you determine dy/dx given sin(xy)=2x+5?

Dec 19, 2016

$y ' = \frac{2 \sec \left(x y\right) - y}{x} , x \in \left[- 3 , - 2\right]$

#### Explanation:

$\sin \left(x y\right) = 2 x + 5 \in \left[- 1. 1\right] \to x \in \left[- 3 , - 2\right]$.

((sin (xy))'=(2x+5)'

cos(xy)(xy'+y)=2#

$y ' = \frac{2 \sec \left(x y\right) - y}{x} , x \in \left[- 3 , - 2\right]$.

Dec 19, 2016

#### Explanation:

One must use the chain rule to differentiate $\sin \left(x y\right)$

$\frac{\mathrm{df} \left(g \left(x , y\right)\right)}{\mathrm{dx}} = \frac{d \left(f \left(g\right)\right)}{\mathrm{dg}} \frac{d \left(g \left(x , y\right)\right)}{\mathrm{dx}}$

let $g \left(x , y\right) = x y$, then $f \left(g\right) = \sin \left(g\right) , \frac{\mathrm{df} \left(g\right)}{\mathrm{dg}} = \cos \left(g\right) ,$ And we need to use the product for the last part:

$\frac{d \left(g \left(x , y\right)\right)}{\mathrm{dx}} = \frac{d \left(x y\right)}{\mathrm{dx}}$

$\frac{d \left(u v\right)}{\mathrm{dx}} = \left(u '\right) \left(v\right) + \left(u\right) \left(v '\right)$

let $u = x$, then $v = y , u ' = 1$ and $v ' = \frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{d \left(g \left(x , y\right)\right)}{\mathrm{dx}} = \left(1\right) \left(y\right) + \left(x\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = y + x \frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{ds} \in \left(x y\right)}{\mathrm{dx}} = \cos \left(x y\right) \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

$\frac{\mathrm{ds} \in \left(x y\right)}{\mathrm{dx}} = y \cos \left(x y\right) + x \cos \left(x y\right) \frac{\mathrm{dy}}{\mathrm{dx}}$

Differentiating the right side is trivial:

$\frac{d \left(2 x + 5\right)}{\mathrm{dx}} = 2$

Put these back into their respective locations within the equation:

$y \cos \left(x y\right) + x \cos \left(x y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 2$

Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

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

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