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Found in: Page 656

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Find the value of $$\frac{{dy}}{{dx}}$$ using equation 6.$$\cos (xy) = 1 + \sin y$$

The value of $$\frac{{dy}}{{dx}}$$ is$$- \frac{{y\sin (xy)}}{{\cos y + x\sin (xy)}}$$.

See the step by step solution

## Step 1: Formula used

Equation 6: "$$\frac{{dy}}{{dx}} = - \frac{{\frac{{\partial F}}{{dx}}}}{{\frac{{\partial F}}{{dy}}}} = - \frac{{{F_x}}}{{{F_y}}}$$, where $$F$$ is the function of $$x$$ and$${y^{\prime \prime }}$$.

## Step 2: Find the value of $$\frac{{dy}}{{dx}}$$

As given equation is $$\cos (xy) = 1 + \sin y$$

Let the function be, $$F(x,y) = \cos (xy) - 1 - \sin y \ldots \ldots \ldots (1)$$

Obtain the equation $$\frac{{dy}}{{dx}}$$ by using the equation 6,

$$\frac{{dy}}{{dx}} = - \frac{{{F_x}}}{{{F_y}}} \ldots \ldots \ldots (2)$$

Take the partial derivative with respect to $$x$$ in the equation (1),

\begin{aligned}{l}\frac{{\partial F}}{{\partial x}} = \frac{\partial }{{\partial x}}(\cos (xy) - 1 - \sin y)\\\frac{{\partial F}}{{\partial x}} = \frac{\partial }{{\partial x}}(\cos (xy)) - \frac{\partial }{{\partial x}}(1) - \frac{\partial }{{\partial x}}(\sin y)\\\frac{{\partial F}}{{\partial x}} = - \sin xy(y) - 0 - 0{F_x}\\\frac{{\partial F}}{{\partial x}} = - y\sin (xy)\end{aligned}

Thus, the partial derivate is $${F_x} = - y\sin (xy)$$.

Take the partial derivative with respect to $$y$$ in the equation (1),

\begin{aligned}{l}\frac{{\partial F}}{{\partial y}} = \frac{\partial }{{\partial y}}(\cos (xy) - 1 - \sin y)\\\frac{{\partial F}}{{\partial y}} = \frac{\partial }{{\partial y}}(\cos (xy)) - \frac{\partial }{{\partial y}}(1) - \frac{\partial }{{\partial y}}(\sin y)\\\frac{{\partial F}}{{\partial y}} = - \sin (xy)(x) - 0 - \cos y{F_y}\\\frac{{\partial F}}{{\partial y}} = - x\sin (xy) - \cos y\end{aligned}

Thus, the partial derivate of$$y,{F_y} = - x\sin (xy) - \cos y$$.

Substitute the respective values in the equation (2),

\begin{aligned}{l}\frac{{dy}}{{dx}} = - \frac{{ - y\sin (xy)}}{{ - x\sin (xy) - \cos y}}\\\frac{{dy}}{{dx}} = - \frac{{y\sin (xy)}}{{x\sin (xy) + \cos y}}\end{aligned}

Thus, the value of $$\frac{{dy}}{{dx}}$$is $$- \frac{{y\sin (xy)}}{{x\sin (xy) + \cos y}}$$.