91Ó°ÊÓ

The Chain Rule states that if we have two functions say \(u(x)\) and \(v(u)\), then the derivative of the composite function \( v(u(x))\) is given by: \[ (v \circ u)' (x) = v'(u(x)) \cdot u'(x) \] This rule allows us to differentiate complex functions by differentiating the inner and outer functions separately.

To prove the chain rule, we will use the definition of a derivative, which states that the derivative of a function \(f(x)\) at a point \(a\) is: \[ f'(a)=\lim_{{h \to 0}} \frac{f(a + h) - f(a)}{h} \] Now, let \(y = v(u(x))\), then the derivative of \(y\) with respect to \(x\) is: \[ y'=\lim _{{h \to 0}} \frac{y(x + h) - y(x)}{h} \] This can be written as: \[ y'=\lim_{{h \to 0}} \frac{v(u(x + h)) - v(u(x))}{h} \] We have a problem here because of \(h\) in the denominator which we can overcome by multiplying and dividing by \(u(x+h) - u(x)\). This gives us: \[ y'=\lim_{{h \to 0}} \frac{v(u(x + h)) - v(u(x))}{u(x+h) - u(x)} \times \frac{u(x+h) - u(x)}{h} \] The first part of the RHS is just the definition of a derivative of \(v(u)\) at \(u(x)\) and the second part of the RHS is the derivative of \(u(x)\). Therefore, we get: \[ y' = v'(u(x)) \cdot u'(x) \] Hence, the Chain Rule is proven.