# 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.

__Proof:__

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**$\dfrac{d}{dx}[c \cdot f(x)]=c \cdot f'(x)$**

__Proof:__

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