# The Sum Rule

Let $ƒ(x)$ and $g(x)$ be any two differentiable functions of $x$. Let $h(x)=ƒ(x)\pm g(x)$ Then, $h'(x)=ƒ'(x)\pm g'(x)$ $\text{or, } \frac{d}{dx}h(x)=\frac{d}{dx}ƒ(x)\pm \frac{d}{dx}g(x)$ Proof: Let $a$ be a fixed point. Then, by first principles, we have, $h'(a)=\lim_{x\to a}\frac{h(x)-h(a)}{x-a}$ $=\lim_{x\to a}\frac{[ƒ(x)\pm g(x)]-[ƒ(a)\pm g(a)]}{x-a}$ $=\lim_{x\to a}\frac{ƒ(x)-ƒ(a)}{x-a}\pm \lim_{x\to a}\frac{g(x)-g(a)}{x-a}$ $=ƒ'(a)\pm g'(a)$ But $a$ is an arbitrary fixed number so $h'(x)=ƒ'(x)\pm g'(x)$ $\frac{d}{dx}[ƒ(x)\pm g(x)]=\frac{d}{dx}ƒ(x)\pm \frac{d}{dx}g(x)$ Put $ƒ(x)=u$ and $g(x)=v$. Then, $\frac{d}{dx}(u\pm v)=\frac{du}{dx}\pm \frac{dv}{dx}$

From this sum rule, we can also deduce a formula for the differentiation of ‘a constant times a function’. $\text{i. }\frac{d(2u)}{dx}=\frac{d(u+u)}{dx}$$=\frac{du}{dx}+\frac{du}{dx}$$=2.\frac{du}{dx}$ $\text{ii. }\frac{d(3u)}{dx}=\frac{(2u)}{dx}+\frac{du}{dx}$ $=2.\frac{du}{dx}+\frac{du}{dx}$ $=3\frac{du}{dx}$ $\text{iii. }\frac{d(4u)}{dx}=\frac{d(3u+u)}{dx}$$=\frac{d(3u)}{dx}+\frac{du}{dx}$ $=4\frac{du}{dx}$ In general, for any integer $n$, $\frac{d(nu)}{dx}=n\frac{du}{dx}$ This law holds for any rational number.

#### Example: Find the derivative of $3x^2-5x+7$.

$\text{Let }y=3x^2-5x+7$ Differentiating both sides w.r.t. $x$, $\frac{dy}{dx}=\frac{d}{dx}(3x^2-5x+7)$ $=\frac{d(3x^2)}{dx}-\frac{d(5x)}{dx}+\frac{d(7)}{dx}$ $=3\frac{dx^2}{dx}-5\frac{dx}{dx}$ $\therefore \frac{dy}{dx}=6x-5$

# The Product Rule

Let $ƒ(x)$ and $g(x)$ be any two differentiable functions of $x$. Let $h(x)=ƒ(x).g(x)$ Then, $h'(x)=ƒ(x).g'(x)+g(x).ƒ'(x)$ $\text{or, }\frac{d}{dx}h(x)=ƒ(x).\frac{d}{dx}g(x)+g(x).\frac{d}{dx}ƒ(x)$ Proof: Let $a$ be a fixed point. Then, by definition, $h'(a)=\lim_{x\to a}\frac{h(x)-h(a)}{x-a}$ $=\lim_{x\to a} \frac{ƒ(x)g(x)-ƒ(a)g(a)}{x-a}$ $=\lim_{x\to a}\frac{ƒ(x)g(x)-ƒ(x)g(a)+ƒ(x)g(a)-ƒ(a)g(a)}{x-a}$ $=\lim_{x\to a}ƒ(x)\lim_{x\to a}\frac{g(x)-g(a)}{x-a}+g(a).\lim_{x\to a}\frac{ƒ(x)-ƒ(a)}{x-a}$ $=ƒ(a)g'(a)+g(a)ƒ'(a)$ But $a$ is an arbitrary fixed number, so $h'(x)=ƒ(x).g'(x)+g(x).ƒ'(x)$ $\frac{d}{dx}[ƒ(x).g(x)]=ƒ(x).\frac{d}{dx}g(x)+g(x).\frac{d}{dx}ƒ(x)$ Put $ƒ(x)=u$ and $g(x)=v$, then, $\frac{d}{dx}(u.v)=u\frac{dv}{dx}+v\frac{du}{dx}$

#### Example: Find the derivative of $(2x^2+1)(3x^2-2)$.

Let $u=2x^2+1$ and $v=3x^2-2$. Then, we have, $\frac{d}{dx}(u.v)=u\frac{dv}{dx}+v\frac{du}{dx}$ $=(2x^2+1)\frac{d}{dx}(3x^2-2)+(3x^2-2)\frac{d}{dx}(2x^2+1)$ $=(2x^2+1)(6x)+(3x^2-2)(4x)$ $=2x(12x^2-1)$

# The Power Rule

Let $u$ be the function of $x$, then $\frac{d}{dx}(u^n)=nu^{n-1}\frac{du}{dx}$ Proof: From the product rule, we have, $\frac{d}{dx}(u^2)=\frac{d}{dx}(u.u)$$=u\frac{du}{dx}+u\frac{du}{dx}=2u\frac{du}{dx}$ Also, $\frac{d}{dx}(u^3)=\frac{d}{dx}(u.u^2)$ $=u\frac{du^2}{dx}+u^2\frac{du}{dx}$ $=u.2u\frac{du}{dx}+u^2\frac{du}{dx}=3u^2\frac{d}{dx}$ So, in general, for any integer $n$, $\frac{d}{dx}(u^n)=nu^{n-1}\frac{du}{dx}$ It holds for any rational number.

#### Example: Find the derivative of $\sqrt{8-5x}$.

$\frac{d}{dx}\sqrt{8-5x}=\frac{d}{dx}(8-5x)^{\frac{1}{2}}$ $=\frac{1}{2}(8-5x)^{-\frac{1}{2}}\frac{d}{dx}(8-5x)$ $=\frac{1}{2\sqrt{8-5x}}×-5$ $=\frac{-5}{2\sqrt{8-5x}}$

# The Quotient Rule

Let $ƒ(x)$ and $g(x)$ be any two differentiable functions of $x$. Let $h(x)=\frac{ƒ(x)}{g(x)}$ Then $h'(x)=\frac{g(x)ƒ'(x)-ƒ(x)g'(x)}{\{g(x)\}^2}$ $\text{or, }\frac{d}{dx}\left\{\frac{ƒ(x)}{g(x)}\right\}=\frac{g(x)\frac{d}{dx}ƒ(x)-ƒ(x)\frac{d}{dx}g(x)}{\{g(x)\}^2}$ Proof: Let $a$ be a fixed point. Then, by first principles, $h'(a)=\lim_{x\to a} \frac{h(x)-h(a)}{x-a}$ $=\lim_{x\to a}\frac{\frac{ƒ(x)}{g(x)}-\frac{ƒ(a)}{g(a)}}{x-a} \text{ [g(a)≠0]}$ $=lim_{x\to a}\frac{g(a)ƒ(x)-ƒ(a)g(x)}{(x-a)g(x)g(a)}$ $=\lim_{x\to a} \frac{g(a)ƒ(x)-g(a)ƒ(a)+g(a)ƒ(a)-ƒ(a)g(x)}{(x-a)g(x)g(a)}$ $=\lim_{x\to a}\left[\frac{1}{g(a)g(x)}\left\{g(a)\frac{ƒ(x)-ƒ(a)}{x-a}-ƒ(a)\frac{g(x)-g(a)}{x-a}\right\}\right]$ $=\frac{1}{g(a)g(a)}[g(a)ƒ'(a)-ƒ'(a)g(a)]$ $=\frac{g(a)ƒ'(a)-ƒ(a)g'(a)}{\{g(a)\}^2}$ But $a$ is an arbitrary fixed number, so $h'(x)=\frac{g(x)ƒ'(x)-ƒ(x)g'(x)}{\{g(x)\}^2}$ $\text{or, }\frac{d}{dx}\left\{\frac{ƒ(x)}{g(x)}\right\}=\frac{g(x)\frac{d}{dx}ƒ(x)-ƒ(x)\frac{d}{dx}g(x)}{\{g(x)\}^2}$ Put $ƒ(x)=u$ and $g(x)=v$. Then, we have, $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$

#### Example: Find the derivative of $\frac{x^2-a^2}{x^2+a^2}$.

$\frac{d}{dx}\left(\frac{x^2-a^2}{x^2+a^2}\right)$ $=\frac{(x^2+a^2)\frac{d}{dx}(x^2-a^2)-(x^2-a^2)\frac{d}{dx}(x^2+a^2)}{(x^2+a^2)^2}$ $=\frac{(x^2+a^2)(2x)-(x^2-a^2)(2x)}{(x^2+a^2)^2}$ $=\frac{2x^3+2a^2x-2x^3+2a^2x}{(x^2+a^2)^2}$ $=\frac{4a^2x}{(x^2+a^2)^2}$

# The Chain Rule

If $y=ƒ(u)$ and $u=g(x)$, where ƒ and g are differentiable functions, then $\frac{dy}{dx}$ exists and $\frac{dy}{dx}=\frac{dy}{du}.\frac{du}{dx}=ƒ'(u)\frac{du}{dx}$ Proof: Let $u=g(x)$. Let $\Delta x$ be a small increment in $x$ and $\Delta u$ be the corresponding small increment in $u$. Then $u+\Delta u=g(x+\Delta x)$ $\Delta u=g(x+\Delta x)-u=g(x+\Delta x)-g(x)$ Since, $g(x)$ is differentiable, it is continuos. Hence, $\lim_{\Delta x\to 0}\Delta u=\lim_{\Delta x\to 0}[g(x+\Delta x)-g(x)]$ $=g(x)-g(x)=0$ Thus $\Delta u\to 0$ as $\Delta x\to 0$. Then $\frac{\Delta y}{\Delta x}=\frac{\Delta y}{\Delta u}.\frac{\Delta u}{\Delta x}$ $\frac{dy}{dx}=\lim_{\Delta x\to 0} \frac{\Delta y}{\Delta x}$ $=\lim_{\Delta x\to 0}\left[\frac{\Delta y}{\Delta u}.\frac{\Delta u}{\Delta x}\right]$ $=\left(\lim_{\Delta u\to 0}\frac{\Delta y}{\Delta u}\right) \left(\lim_{\Delta x\to 0}\frac{\Delta u}{\Delta x}\right)$ $=\frac{dy}{du}.\frac{du}{dx}$

#### Example: Calculate $\frac{dy}{dx}$ is $y=5t^2+6t-7$ and $t=x^3-2$.

$\frac{dy}{dx}=\frac{dy}{dt}.\frac{dt}{dx}$ $=\frac{d}{dt}(5t^2+6t-7).\frac{d}{dx}(x^3-2)$ $=(10t+6)(3x^2)$ $=6x^2(5t+3)$ $=6x^2[5(x^3-2)+3]$ $=6x^2(5x^3-7)$

# Second and Higher Derivatives

Let $y=ƒ(x)$ be a differentiable function. Then $\frac{dy}{dx}=ƒ'(x)$ is called the first derivative or the first differential coefficient of $ƒ(x)$ with respect to $x$. If we differentiate $\frac{dy}{dx}$ again, we get $\frac{d}{dx}\left(\frac{dy}{dx}\right)$ written as $\frac{d^2y}{dx^2}$ or $ƒ”(x)$, which is called the second derivative or the second differential coefficient of $ƒ(x)$ with respect to $x$ and so on.

#### Example: Find the second and higher derivatives of $y=3x^4-x^2+1$.

$\begin{array}{c} \frac{dy}{dx}=12x^3-2x, & \text{ the first derivative} \\ \frac{d^2y}{dx^2}=36x^2-2, & \text{ the second derivative} \\ \frac{d^3y}{dx^3}=72x, & \text{ the third derivative} \\ \frac{d^4y}{dx^4}=72, & \text{ the fourth derivative} \\ \frac{d^5y}{dy^5}=0, & \text{ the fifth derivative} \end{array}$ and all the other derivatives are also zero.

# Implicit Function and Implicit Differentiation

Let $ƒ(x, y)$ be an arbitrary function of two variables $x$ and $y$ and let $ƒ(x,y)=0$. This equation may or may not be solvable for $y$. But it can be differentiated term by term and $\frac{dy}{dx}$ can be solved. This process of finding the value of $\frac{dy}{dx}$ without solving the equation for $y$ is called implicit differentiation.

#### Example: Use implicit differentiation to find $\frac{dy}{dx}$ in $x^2y^2=x^2+y^2$.

We have, $x^2y^2=x^2+y^2$ Differentiating with respect to $x$, $\frac{d}{dx}(x^2y^2)=\frac{d}{dx}(x^2+y^2)$ $x^2\frac{dy^2}{dx}+y^2\frac{dx^2}{dx}=2x+2y\frac{dy}{dx}$ $x^2.2y\frac{dy}{dx}+y^2.2x=2x+2y\frac{dy}{dx}$ $2x^2y\frac{dy}{dx}-2y\frac{dy}{dx}=2x-2xy^2$ $\frac{dy}{dx}=\frac{2x-2xy^2}{2x^2y-2y}$ $\therefore \frac{dy}{dx}=\frac{x(1-y^2)}{y(x^2-1)}$