For a function of several variables like f(x,y), a partial derivative with respect to one variable treats every other variable as a constant:
∂x∂f=h→0limhf(x+h,y)−f(x,y)
Worked example: first partial derivatives
For f(x,y)=x3y2+sin(xy):
∂x∂f=3x2y2+ycos(xy),∂y∂f=2x3y+xcos(xy)
The y in ycos(xy) comes from the chain rule: the derivative of xy with respect to x is y.
Worked example: mixed partials agree
Differentiating ∂x∂f with respect to y, and ∂y∂f with respect to x, both give:
∂y∂x∂2f=∂x∂y∂2f=6x2y+cos(xy)−xysin(xy)
Worked example: gradient and directional derivative
For g(x,y)=x2y+y3 at (1,2): gx=2xy=4 and gy=x2+3y2=13, so ∇g=(4,13). The rate of change in the direction of (3,4), using the unit vector (53,54), is:
Du^g=4(53)+13(54)=564=12.8
The largest possible rate of change is ∣∇g∣=16+169=185≈13.6, in the direction of ∇g itself.
Worked example: a minimum of two variables
Find the extreme value of h(x,y)=x2+y2−2x−4y.
Solution:hx=2x−2=0 and hy=2y−4=0 give the critical point (1,2). With r=hxx=2, s=hxy=0 and t=hyy=2, rt−s2=4>0 and r>0, so it's a minimum: h(1,2)=1+4−2−8=−5.
Common mistakes
Treating the other variable as a function. In ∂/∂x, y is a constant: ∂(x2y)/∂x=2xy.
Forgetting the chain-rule factor in terms like sin(xy).
Using a non-unit direction vector. Divide by its length before taking the directional derivative.
Calling every critical point an extremum. If rt−s2<0 it's a saddle point; if it's 0 the test is inconclusive.
Where this comes up in exams
Partial derivatives and maxima and minima are listed under Calculus in the GATE 2027 EC, EE and IN syllabi, and the EE syllabus also lists directional derivatives.