Techniques of Differentiation

Some techniques of differentiation are listed below:

  1. The Sum Rule
  2. The Product Rule
  3. The Power Rule
  4. The Quotient Rule
  5. The Chain Rule
  6. Second and Higher Derivatives
  7. Implicit Function and Implicit Differentiation

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 $x$, \[\frac{d}{dx}(x^2y^2)=\frac{d}{dx}(x^2+y^2)\] \[x^2\frac{dy^2}{dx}+y^2\frac{dx^2}{dx}=2x+2y\frac{dy}{dx}\] \[x^2.2y\frac{dy}{dx}+y^2.2x=2x+2y\frac{dy}{dx}\] \[2x^2y\frac{dy}{dx}-2y\frac{dy}{dx}=2x-2xy^2\] \[\frac{dy}{dx}=\frac{2x-2xy^2}{2x^2y-2y}\] \[\therefore \frac{dy}{dx}=\frac{x(1-y^2)}{y(x^2-1)}\]


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