Question

find the first partial derivates of the function

\(U = xy{\sin ^ - }\left( {yz} \right)\)A

Short answer

The first partial derivates are \({U_x} = xy{\sin ^ - }\left( {yz} \right)\)

\({U_y} = \frac{{xyz}}{{\sqrt {1 - {z^2}{y^2}} }} + x{\sin ^ - }\left( {yz} \right)\) and \({U_z} = \frac{{x{y^2}}}{{\sqrt {1 - {y^2}{z^2}} }}\)

Step-by-step solution

Step1:- finding the first partial derivates with respect to x.

\({U_x} = xy{\sin ^ - }\left( {yz} \right)\)

Differentiating its partially with respect to x

\(\begin{aligned}{l}{U_x} &= \frac{\partial }{{\partial x}}(xy{\sin ^ - }\left( {yz} \right))\\{U_x} &= y{\sin ^ - }\left( {yz} \right)\frac{\partial }{{\partial x}}(x)\\{U_x} &= y{\sin ^ - }\left( {yz} \right)\end{aligned}\)

Step2:- finding the partial derivative with respect to y.

\(U = xy{\sin ^ - }\left( {yz} \right)\)

Differentiating it partial with respect to y

\(\begin{aligned}{l}{U_y} &= \frac{\partial }{{\partial y}}(xy{\sin ^ - }\left( {yz} \right))\\{U_y} &= x(y{\sin ^ - }\left( {yz} \right) + {\sin ^ - }\left( {yz} \right)\frac{\partial }{{\partial y}}(y))\\{U_y} &= x(y.\frac{1}{{\sqrt {1 - {y^2}{z^2}} }}\frac{\partial }{{\partial y}}\left( {yz} \right) + {\sin ^ - }\left( {yz} \right))\\{U_y} &= x(\frac{y}{{\sqrt {1 - {y^2}{z^2}} }}.z + {\sin ^ - }\left( {yz} \right))\\{U_y} &= \frac{{xyz}}{{\sqrt {1 - {y^2}{z^2}} }} + x{\sin ^ - }\left( {yz} \right)\end{aligned}\)

Step3:- finding the first partial derivative with respect of z.

\(U = xy{\sin ^ - }\left( {yz} \right)\)

Differentiating it partial with respect to z

\(\begin{aligned}{l}{U_z} &= \frac{\partial }{{\partial z}}(xy{\sin ^ - }(yz))\\{U_z} &= xy\frac{\partial }{{\partial z}}\left( {{{\sin }^ - }(yz)} \right)\\{U_z} &= xy\frac{1}{{\sqrt {1 - {y^2}{z^2}} }}\frac{\partial }{{\partial z}}\left( {yz} \right)\\{U_z} &= \frac{{xy}}{{\sqrt {1 - {y^2}{z^2}} }}.y\\{U_z} &= \frac{{x{y^2}}}{{\sqrt {1 - {y^2}{z^2}} }}\end{aligned}\)

Hence, the partial derivatives are \({U_x} = xy{\sin ^ - }\left( {yz} \right)\)

\({U_y} = \frac{{xyz}}{{\sqrt {1 - {z^2}{y^2}} }} + x{\sin ^ - }\left( {yz} \right)\) and \({U_z} = \frac{{x{y^2}}}{{\sqrt {1 - {y^2}{z^2}} }}\)