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3.18 Proof of the Chain Rule 1

Rough proof (not examinable). Let x=x⁢(t)x=x(t) and y=y⁢(t)y=y(t). Suppose we make a small change ss in the value of tt. Let h=x⁢(t+s)-x⁢(t)h=x(t+s)-x(t) and k=y⁢(t+s)-y⁢(t)k=y(t+s)-y(t). Then hs→d⁢xd⁢t\frac{h}{s}\rightarrow\frac{dx}{dt} and ks→d⁢yd⁢t\frac{k}{s}\rightarrow\frac{dy}{dt} as s→0s\rightarrow 0. Thus

hs⁢∂⁡f∂⁡x+ks⁢∂⁡f∂⁡y→d⁢xd⁢t⁢∂⁡f∂⁡x+d⁢yd⁢t⁢∂⁡f∂⁡y⁢as s→0s\rightarrow 0.\frac{h}{s}\frac{\partial f}{\partial x}+\frac{k}{s}\frac{\partial f}{\partial y% }\rightarrow\frac{dx}{dt}\frac{\partial f}{\partial x}+\frac{dy}{dt}\frac{% \partial f}{\partial y}\;\;\mbox{as $s\rightarrow 0$.}

Now f⁢(x⁢(t+s),y⁢(t+s))-f⁢(x⁢(t+s),y)s=\frac{f(x(t+s),y(t+s))-f(x(t+s),y)}{s}=

f⁢(x+h,y+k)-f⁢(x+h,y)k⋅ks→∂⁡f∂⁡y⋅d⁢yd⁢t\frac{f(x+h,y+k)-f(x+h,y)}{k}\cdot\frac{k}{s}\rightarrow\frac{\partial f}{% \partial y}\cdot\frac{dy}{dt}

as s→0s\rightarrow 0. Similarly, f⁢(x⁢(t+s),y)-f⁢(x,y)s=\frac{f(x(t+s),y)-f(x,y)}{s}=

f⁢(x+h,y)-f⁢(x,y)h⋅hs→∂⁡f∂⁡x⋅d⁢xd⁢t⁢as s→0s\rightarrow 0.\frac{f(x+h,y)-f(x,y)}{h}\cdot\frac{h}{s}\rightarrow\frac{\partial f}{\partial x% }\cdot\frac{dx}{dt}\;\;\mbox{as $s\rightarrow 0$}.

Summing the two expressions, we obtain the required value for the derivative.