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Q22E

Expert-verifiedFound in: Page 656

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\( - \frac{{y\sin (xy)}}{{\cos y + x\sin (xy)}}\).

**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**\({y^{\prime \prime }}\)**.**

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 \( - \frac{{y\sin (xy)}}{{x\sin (xy) + \cos y}}\).

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