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partial_derivatives

# partial_derivatives - x I distinguish the partial...

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Partial Deriviatives in PHYS171 ALP September 3, 2011 In lecture I introduced a concept that was unfamiliar to many of you. This is a partial derivative . Partial deriviatives often come into play in very complicated math- ematical situations, however the idea itself is simple. The simplest case in calculus is a derivative of a function of one variable. If f ( x ) = x 3 df dx = d dx x 3 = 3 x 2 . What if we have a function of two variables, f ( x, y ) = x 3 + y 4 and we want to take the deriviative with respect to x ? We would write df dx = d dx ( x 3 + y 4 ) = d dx x 3 + d dx y 4 = 3 x 2 + 4 y 3 dy dx where we have used the chain rule to evaluate the term containing y . This is called the complete derivative, we are taking account of the possibility that y depends on x and that the change in y with x will contribute to the change in f . The concept I used when discussing error analysis was the partial derivative. This means I take the derivative of the function with respect to x , assuming that all other variables are constant (ignoring the possibility that other variables depend on
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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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