Engineering Calculus Notes 247

Engineering Calculus Notes 247 - single-variable calculus...

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3.3. DERIVATIVES 235 while the partial with respect to y is found by treating x as a constant: f y ( x,y ) := ∂f ∂y ( x,y ) = x 2 + 4; similarly, if g ( x,y,z ) = sin 2 x cos y + xyz 2 then g x ( x,y,z ) := ∂g ∂x ( x,y,z ) = 2 cos 2 x cos y + yz 2 g y ( x,y,z ) := ∂g ∂y ( x,y,z ) = sin 2 x sin y + xz 2 g z ( x,y,z ) := ∂g ∂z ( x,y,z ) = 2 xyz. Remark 3.2.1 tells us that the differential of f , being linear, is determined by the partials of f : d −→ x 0 f ( −→ v ) = parenleftbigg ∂f ∂x 1 ( −→ x 0 ) parenrightbigg v 1 + parenleftbigg ∂f ∂x 2 ( −→ x 0 ) parenrightbigg v 2 + parenleftbigg ∂f ∂x 3 ( −→ x 0 ) parenrightbigg v 3 = 3 summationdisplay j =1 ∂f ∂x j v j . So far, we have avoided the issue of existence: all our formulas above assume that f ( −→ x ) is differentiable at −→ x = −→ x 0 . Since the partial derivatives of a function are essentially derivatives as we know them from
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Unformatted text preview: single-variable calculus, it is usually pretty easy to determine whether they exist and if so to calculate them formally. However, the existence of the partials is not by itself a guarantee that the function is diferentiable . ±or example, the function we considered in § 3.1 f ( x ) = b xy x 2 + y 2 if ( x,y ) n = (0 , 0) , at (0 , 0) has the constant value zero along both axes, so certainly its two partials at the origin exist and equal zero ∂f ∂x (0 , 0) = 0 ∂f ∂y (0 , 0) = 0...
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