AMH - 28__calc.28_prop

Claim: If \(c\) is a contant and \(f\) is a differentiable function at \(a,\) then \[(cf)'(a) = cf'(a)\]

▼ Proof:

By definition, the derivative of \(cf\) at \(a\) is \[(cf)'(a) = \lim_{h \rightarrow 0}\frac{(cf)(a+h) - (cf)(a)}{h}\] Pulling out the constant leaves you \[c\lim_{h \rightarrow 0}\frac{f(a+h) - f(a)}{h} = cf'(a)\]

Example: The derivative of \(x^2\) is \(2x.\) So, the derivative of \(5x^2\) is \(5\) times the derivative of \(x^2,\) or \begin{align} (5x^2)' & = 5(x^2)' \\ & = 5(2x) \\ & = 10x \end{align}

Claim: If \(f\) and \(g\) are differentiable functions at \(a,\) then \[(f+g)'(a) = f'(a)+g'(a)\]

▼ Proof:

The proof follows from the definition of a derivative and properties of limits: \begin{align} (f+g)'(a) & = \lim_{h \rightarrow 0} \frac{(f+g)(a+h) - (f+g)(a)}{h} \\ & = \lim_{h \rightarrow 0} \frac{f(a+h)+g(a+h) - f(a)-g(a)}{h} \\ & = \lim_{h \rightarrow 0} \frac{f(a+h)- f(a)+g(a+h) -g(a)}{h} \\ & = \lim_{h \rightarrow 0} \frac{f(a+h)- f(a)}{h}+\lim_{h \rightarrow 0}\frac{g(a+h) -g(a)}{h} \\ & = f'(a)+g'(a) \end{align}

Example: The derivative of \(x^2\) is \(2x\) and the derivative of \(x^3\) is \(3x^2.\) So, \begin{align} (x^2+x^3)' & = (x^2)' + (x^3)' \\ & = 2x + 3x^2 \end{align}

Corollary: If \(f\) and \(g\) are differentiable functions at \(a\) and \(c\) and \(d\) are constants, then \[(cf(a)+dg(a))' = cf'(a)+dg'(a)\]

Example: By the corollary, \begin{align} (7x^2+5x^3)' & = (7x^2)' + (5x^3)' \\ & = 7(x^2)' + 5(x^3)' \\ & = 7(2x) + 5(3x^2) \\ & = 14x + 15x^2 \end{align}