Proofs of Derivative Properties

Properties of Derivatives and Their Proofs

In this section, we will prove various properties of derivatives with applications. The following notation will be used to understand the derivative of $f(x)$: $f'(x)=\dfrac{d}{dx}(f(x))$.


Theorem 1:  Prove that the derivative of a constant function is zero.


Let $f(x)=c$ be a constant function. We have to show that $\dfrac{d}{dx}(c)=0$, that is, $\dfrac{d}{dx}(f(x))=0$.

Now, by definition of derivative we have

$\dfrac{d}{dx}(c)=\dfrac{d}{dx}(f(x))=$ $\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$ $=\lim\limits_{h \to 0} \dfrac{c-c}{h}$ $=\lim\limits_{h \to 0} \dfrac{0}{h}$ $=0$ ans.


Theorem 2:  Let  $c$ be a constant and $f(x)$ be a differentiable function. Then we have
$\dfrac{d}{dx}[c \cdot f(x)]=c \cdot f'(x)$
Let $g(x)=c f(x)$.
From the definition of the derivative, we have
$g'(x)=\dfrac{d}{dx}(g(x))=$ $\lim\limits_{h \to 0} \dfrac{g(x+h)-g(x)}{h}$

$=\lim\limits_{h \to 0} \dfrac{c f(x+h)-c f(x)}{h}$

$=c\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$

$=c f'(x)$ (by definition) ans.


Let $u$ and $v$ be two differentiable functions of $x$.

Theorem 3:  Sum Rule of Derivatives: $\dfrac{d}{dx}(u+v)=\dfrac{du}{dx}+\dfrac{dv}{dx}$

Proof: Let $f(x)=u(x)+v(x)$.

By the definition of differentiation, we get that
$f'(x)=\dfrac{d}{dx}(f(x))=$ $\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$

$=\lim\limits_{h \to 0}$ $\dfrac{u(x+h)+v(x+h)-u(x)-v(x)}{h}$ 

$=\lim\limits_{h \to 0}$ $\dfrac{[u(x+h)-u(x)]+[v(x+h)-v(x)]}{h}$

$=\lim\limits_{h \to 0}$ $\dfrac{u(x+h)-u(x)}{h}+\dfrac{v(x+h)-v(x)}{h}$

$=\lim\limits_{h \to 0} \dfrac{u(x+h)-u(x)}{h} +$ $\lim\limits_{h \to 0} \dfrac{v(x+h)-v(x)}{h}$

$=u'(x)+v'(x)$ (by definition)

Therefore, $\dfrac{d}{dx}(u+v)=\dfrac{du}{dx} + \dfrac{dv}{dx}$


In a similar way, one can prove the Difference Rule of Derivatives: $\dfrac{d}{dx}(u-v)=\dfrac{du}{dx}-\dfrac{dv}{dx}$


Theorem 4:  Let 


Theorem 5:  Let  
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