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2.4 Partial differentiation from first principles

Example.

To find the partial derivatives of f⁢(x,y)=x2+x⁢yf(x,y)=x^{2}+xy.

Solution.

Let (x,y)=(a,b)(x,y)=(a,b). To determine ∂⁡f∂⁡x\frac{\partial f}{\partial x} we need to calculate the change that occurs in the value of ff for a small change hh in the value of xx. Thus we consider f⁢(a+h,b)=(a+h)2+(a+h)⁢b=a2+2⁢a⁢h+h2+a⁢b+b⁢h.f(a+h,b)=(a+h)^{2}+(a+h)b=\,{a^{2}+2ah+h^{2}+ab+bh.} Then

∂⁡f∂⁡x=limh→0⁡f⁢(a,b+h)-f⁢(a,b)h=limh→0⁡(2⁢a+b+h)=2⁢a+b.\frac{\partial f}{\partial x}=\lim_{h\rightarrow 0}\frac{f(a,b+h)-f(a,b)}{h}=% \,{\lim_{h\rightarrow 0}(2a+b+h)}\,{=2a+b.}

Similarly, f⁢(a,b+k)-f⁢(a,b)=a2+a⁢(b+k)-a2-a⁢b=a⁢k.f(a,b+k)-f(a,b)=a^{2}+a(b+k)-a^{2}-ab=\,{ak.} Therefore

∂⁡f∂⁡y=limk→0⁡f⁢(a,b+k)-f⁢(a,b)k=a.\frac{\partial f}{\partial y}=\lim_{k\rightarrow 0}\frac{f(a,b+k)-f(a,b)}{k}={% a.}