Question

Calc III Tutorial: Partial Derivatives

Answer 1

\({\bf{Definition:}}\)

\(\frac{ ∂f }{ ∂x } = \lim_{h \rightarrow 0}\frac{ f(x0 + h, y0) - f(x0,y0) }{ h }\)
gives the partial derivative at a point (x0,y0)

similar to the derivative definition for single-variable derivatives except this time you are introducing a new variable that essentially acts as a constant along another axis
if you wanted to find the partial derivative wrt y, you would simply take

\(\frac{ ∂f }{ ∂y } = \lim_{h \rightarrow 0}\frac{ f(x0, y0+h) - f(x0,y0) }{ h }\)

visually you can think of it as taking the tangent to a surface on one axis only

Answer 2

\({\bf{How~to~Take~A~Partial~Derivative:}}\)
every independent variable, with the exception of the one you are taking the derivative to, should be treated as a constant

ex. find ∂f/∂y for f(x) = ysin(xy)
treating x as a constant, use the product rule and chain rule for derivatives

= ycos(xy) * ∂/dy (xy) + sin(xy)
= xycos(xy) + sin(xy)

for multi-variable functions try to write out the intermediate steps so you don't skip any derivatives

ex. f(x,y,z) = xsin(y + 3z), find ∂y/∂z
using chain rule and product rule again,
= xcos(y+3x)∂/dz(y + 3z) = xcos(y+3x)(3)

Answer 3

\({\bf{Second~Order~Partial~Derivatives:}}\)

\(\frac{ ∂^{2}f }{ ∂x^{2} }\) = ∂/∂x(∂f/∂x)
\(\frac{ ∂^{2}f }{ ∂x∂y }\) = ∂/∂x(∂f/∂y)
with mixed derivatives (where you are taking the derivative wrt different variables) go from right to left. so with \(\frac{ ∂^{2}f }{ ∂x∂y }\) you would take the derivative wrt y first, then after that, take the derivative of the result wrt x

Clairaut's Theorem: with mixed derivatives \[f _{xy}(a,b) = f _{yx}(a,b)\]
*** however, this only applies if **all** partial derivatives involved are defined on an open region containing (a,b) and are continuous at (a,b) ***

Answer 4

\({\bf{Differentiability:}}\)
f(x,y) is differentiable at (x0,y0) if the partial derivatives of (x0,y0) wrt to both x and y both exist and

\[Δz = f(x0,y0)Δx + f(x0,y0)Δy + \epsilon_{1}Δx + \epsilon_{2}Δy\]
as both (epsilon1, epsilon2) and (x0,y0) approach 0

if it satisfies this condition through its whole domain it's differentiable (smooth surface)

** unlike with single-variable functions, differentiability does not imply continuity, with one condition: if the partial derivatives of f(x,y) exist and are continuous on a disc centered at (x0,y0) then the function is continuous **

as you might expect, continuity through an open region does imply differentiability

Answer 5

Source material is section 14.3 of Thomas' Calculus, Early Transcendentals, 14th edition by Hass, Heil, Weir, et. al.