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# 2. Properties of Derivatives

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}