Let's find the partial derivatives:
∂ f ∂ x = ∂ ∂ x ( x y ) = 1 y ⇒ ∂ f ∂ x ( P ) = 1 \frac{{\partial f}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {\frac{x}{y}} \right) = \frac{1}{y} \Rightarrow \frac{{\partial f}}{{\partial x}}\left( P \right) = 1 ∂ x ∂ f = ∂ x ∂ ( y x ) = y 1 ⇒ ∂ x ∂ f ( P ) = 1
∂ f ∂ y = ∂ ∂ y ( x y ) = − x y 2 ⇒ ∂ f ∂ y ( P ) = − 1 1 2 = − 1 \frac{{\partial f}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {\frac{x}{y}} \right) = - \frac{x}{{{y^2}}} \Rightarrow \frac{{\partial f}}{{\partial y}}\left( P \right) = - \frac{1}{{{1^2}}} = - 1 ∂ y ∂ f = ∂ y ∂ ( y x ) = − y 2 x ⇒ ∂ y ∂ f ( P ) = − 1 2 1 = − 1
find the direction cosines:
∣ v ‾ ∣ = 0 2 + ( − 1 ) 2 = 1 \left| {\overline v } \right| = \sqrt {{0^2} + {{( - 1)}^2}} = 1 ∣ v ∣ = 0 2 + ( − 1 ) 2 = 1
cos α = 0 1 = 0 , cos β = − 1 1 = − 1 \cos \alpha = \frac{0}{1} = 0,\,\,\cos \beta = - \frac{1}{1} = - 1 cos α = 1 0 = 0 , cos β = − 1 1 = − 1
Then
d f d v = ∂ f ∂ x ( P ) cos α + ∂ f ∂ y ( P ) cos β = 1 ⋅ 0 − 1 ⋅ ( − 1 ) = 1 \frac{{df}}{{dv}} = \frac{{\partial f}}{{\partial x}}\left( P \right)\cos \alpha + \frac{{\partial f}}{{\partial y}}\left( P \right)\cos \beta = 1 \cdot 0 - 1 \cdot \left( { - 1} \right) = 1 d v df = ∂ x ∂ f ( P ) cos α + ∂ y ∂ f ( P ) cos β = 1 ⋅ 0 − 1 ⋅ ( − 1 ) = 1
Answer: 1
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