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Unformatted text preview: x ). I distinguish the partial derivative by writing a script d instead of a normal d. (I heard after lecture that my script d looks a lot like the number 2.) ∂f ∂x = ∂ ∂x ( x 3 + y 4 ) = ∂ ∂x x 3 + ∂ ∂x y 4 = 3 x 2 The difference is that ∂ ∂x y 4 = 0 because we are assuming y is simply a constant, independent of x . It equals zero just as ∂ ∂x 7 4 = 0 Suppose we have f ( x, y ) = x 3 y 4 . df dx = d dx ( x 3 y 4 ) = 2 xy 4 + x 2 4 y 3 dy dx where we have used the product rule. The partial derivative is simply ∂f ∂x = ∂ ∂x ( x 3 y 4 ) = 2 xy 4 . We simply treat y as a constant. If this seems odd, consider the formula for the area of a circle, A = πr 2 . If we take the derivative with respect to r we get dA dr = 2 πr . π is just a constant so no need to be concerned with dπ dr . 1...
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This note was uploaded on 12/29/2011 for the course PHYSICS 171 taught by Professor La porta during the Fall '11 term at Maryland.
- Fall '11
- LA PORTA