Chain Rule, Product Rule and Quotient rule proofs

Proof of the product rule

Since $f$ and $g$ are differentiable, there exists functions $F_a,G_a$ which are continuous at a such that $f(x)=f(a)+F_a(x)(x-a)$ and $g(x)=g(a)+G_a(x)(x-a)$. Hence $f(x)g(x)=f(a)g(a)+(F_a(x)g(x)+G_a(x)f(x)+F_a(x)G_a(x)(x-a))(x-a)$.

By the algebra of continuous functions, $F_{a}(x)g(x)+G_{a}(x)f(x)+F_{a}(x)G_{a}(x)(x-a)$ is continuous at a. Its value at a is$F_{a}(x)g(x)+G_{a}(x)f(x)+0$

$=f’(a)g(a)+g’(a)f(a)$.

This is an equivalent definition of continuity.

Proof of the quotient rule?

NOTE FROM MFA: If $g(a)\ne 0$ and g is continuous at $a$ then $g(x)\ne 0$ for some open neighbourhood of a.

When calculating the limit for $x→a$ we will assume $x$ is in the open neighbourhood of $a$ as above.$\left(\frac{1}{g}\right)^{\prime}\left(a\right)=\lim_{x\to a}\frac{\frac{1}{g\left(x\right)}-\frac{1}{g\left(a\right)}}{x-a}=\lim_{x\to a}\frac{-1}{g\left(x\right)g\left(a\right)}\frac{g\left(x\right)-g\left(a\right)}{x-a}=-\frac{1}{g\left(a\right)^2}g^{\prime}\left(a\right)=\frac{g^{\prime}\left(a\right)}{g\left(a\right)^2}$ . Now apply product rule to $f$ and $1/g$.

Proof of the chain rule?

By results in the course, $f(x)-f(\ell)=F_\ell(x)(x-\ell)$ where $F_\ell$ is continuous at $\ell=g(k)$.

Taking $x=g(y)$ gives $f(g(y))-f(g(k))=F_\ell(g(y))(g(y)-g(k)).$

Since $g$ is differentiable at $k$, by results in the course $g(y)-g(k)=G_k(y)(y-k)$ where $G_k$ is continuous at $k$.

Hence $f(g(y))-f(g(k))=F_\ell(g(y))G_k(y)(y-k).$

The function $F_\ell(g(y))G_k(y)$ is continuous at $k$ (composition of cts functions and Algebra of cts functions), so by results in the course. $f\circ g$ is differentiable at $k$ with derivative $F_\ell(g(k))G_k(k)=f'(g(k))g'(k)$.

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