Methods of Differentiation

Differentiation is a fundamental concept in calculus that deals with how things change. Specifically, it's about understanding the rate at which a quantity changes with respect to another quantity. 

In the world of Math, differentiation is like a superpower. It helps us understand how things change over time. Imagine you're driving a car and want to know how fast you're going at any moment. That's where differentiation comes in. It's like a magic trick that tells us about the speed at any point of our journey. But it's not just for cars; it helps us understand everything from how populations grow to how temperatures fluctuate etc. So, let's dive into this cool math tool and see how it works!

1.0What is Differentiation?

Differentiation is a core concept in calculus that involves finding the rate at which a function changes with respect to its variable. It determines the derivative, which represents the slope of the function at any given point, indicating how the function's value changes as the input changes.

2.0What are Methods of Differentiation?

Differentiation is the process of determining a function's derivative, typically representing a dependent variable in terms of independent variables through an equation. In the following sections, we'll delve into various differentiation methods, offering thorough explanations and relevant formulas for each approach.

• Differentiation using Chain Rule
• Differentiation using Product Rule
• Differentiation using Quotient Rule
• Differentiation through Logarithm
• Differentiation of Parametric Functions
• Differentiation of Implicit Functions

3.0Differentiation using Chain Rule

The Chain Rule serves to differentiate composite functions, stipulating that the derivative of such a composite function equals the product of the outer function's derivative with the inner function's derivative. Mathematically, it's expressed as $\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$. This rule is essential for finding derivatives of functions involving nested functions and plays a crucial role in calculus.

4.0Differentiation using Product Rule

The Product Rule is a fundamental technique in calculus used to differentiate the product of two functions. It allows us to find the derivative of a function that is the product of two simpler functions. Here's how it works:

Let's say we have a function f(x) that is the product of two functions, u(x) and v(x). The Product Rule specifies that the derivative of f(x) with respect to x is:

$\frac{d}{dx}[u(x)\cdot v(x)]=u^{\prime }(x)\cdot v(x)+u(x)\cdot v^{\prime }(x)$

5.0Differentiation using Quotient Rule

The Quotient Rule is a fundamental technique in calculus used to differentiate a quotient of two functions. It allows us to find the derivative of a function that is the ratio of two functions. Here's how it works:

Let's say we have a function f(x) that is the quotient of two functions, u(x) and v(x), where v(x) is not equal to zero. The Quotient Rule specifies that the derivative of f(x) with respect to x is:

$\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right]=\frac{u^{\prime }(x)\cdot v(x)-u(x)\cdot v^{\prime }(x)}{(v(x){)}^2}$

6.0Differentiation through Logarithm

Differentiation using logarithms is a technique employed when dealing with functions containing logarithmic terms. The natural logarithm, denoted as ln(x), is often used in differentiation due to its useful properties. Here's how it works:

Let's say we have a function f(x) that contains a logarithmic term, such as ln(g(x)). To find the differentiation of this function with respect to x, we can use the chain rule, as follows:

$\frac{d}{dx}[\ln (g(x))]=\frac{1}{g(x)}\cdot \frac{dg(x)}{dx}$

7.0Differentiation of Parametric Functions

Differentiating parametric functions involves finding the derivatives of functions defined by parametric equations. These equations express the coordinates of a point on a curve in terms of an independent parameter, usually denoted as t.  Here's how it works:

Let's say we have parametric equations x = f(t) and y = g(t) that define the coordinate (x, y) of a point on a curve.

1. Find the Derivative of x w.r.t t: Differentiate the function x = f(t) with respect to t to find $\frac{dx}{dt}$ which represents the rate of change of x with respect to t.
2. Find the Derivative of y w.r.t t: Differentiate the function y = g(t) with respect to t to find $\frac{dy}{dt}$ which represents the rate of change of y with respect to t.
3. Use the Chain Rule to Find $\frac{dy}{dx}$ : Once you have $\frac{dx}{dt}$ and $\frac{dy}{dt}$, you can find the derivative $\frac{dy}{dx}$ using the chain rule:$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
4. Interpret the Result: The derivative $\frac{dy}{dx}$ represents the slope of the tangent line to the curve at any point. It describes how y changes with respect to x as the parameter t changes.

Parametric differentiation is commonly used in physics, engineering, and geometry to describe the motion of objects, curves in the plane, and various other phenomena. It enables us to analyze and understand the behavior of dynamic systems and complex curves in terms of a single parameter.

8.0Differentiation of Implicit Functions

$\varphi (x,y)=0$

(a) To find $\frac{dy}{dx}$ of implicit functions, we differentiate each term w.r.t. regarding y as a function ofthen collect terms with $\frac{dy}{dx}$ together on one side.

(b) Typically, with implicit functions, both $x\&y$ are present in answers of $\frac{dy}{dx}$. Corresponding to every curve represented by an implicit equation, there exist one or more explicit functions representing that equation. It can be shown that $\frac{dy}{dx}$ at any point on the curve remains the same whether the process of differentiation is done explicitly or implicitly.

9.0Solved Examples on Differentiation

Example1: Find derivative by using 1^st principle of $f(x)=\sin x\text{. }$

Solution:

$f^{\prime }(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}={\lim }_{h\to 0}\frac{\sin (x+h)-\sin x}{h}={\lim }_{h\to 0}\frac{2\cos \left(x+\frac{h}{2}\right)\sin \left(\frac{h}{2}\right)}{h}$

$={\lim }_{h\to 0}\frac{\cos \left(x+\frac{h}{2}\right)\cdot \sin \left(\frac{h}{2}\right)}{h/2}$

$=\cos x\cdot 1=\cos x$

Example 2: If $y=x^x$ , then find $\frac{dy}{dx}$

Solution:

Method 1:

$\log y=\log x^x$

$\Rightarrow \log y=x\log x$

$\Rightarrow \left[\frac{1}{y}\cdot \frac{dy}{dx}\right]=\left[1\cdot \log x+x\frac{1}{x}\right]$

$\Rightarrow \frac{dy}{dx}=y[\log x+1]$

$\Rightarrow \frac{dy}{dx}=x^x[\log x+1]$

Method 2:

$y=e^{x\ln x}$

$\Rightarrow \frac{dy}{dx}=e^{x\ln x}\left(\frac{d}{dx}x\ln x\right)$

$\Rightarrow x^x\left[x\cdot \frac{1}{x}+\ln x\right]=x^x[1+\ln x]$

Example 3: If $y=(\sin x{)}^{\cos x}$, then find $\frac{dy}{dx}$

Solution:

Method 1: $\log y=\log (\sin x{)}^{\cos x}$

$\Rightarrow \log y=\cos x\cdot \log (\sin x)$

$\Rightarrow \frac{1}{y}\cdot \frac{dy}{dx}=\left[-\sin x\cdot \log (\sin x)+\frac{\cos x\cdot 1\cdot \cos x}{\sin x}\right]$

$\Rightarrow \frac{dy}{dx}=y\left[-\sin x\cdot \log (\sin x)+\frac{{\cos }^{2}x}{\sin x}\right]$

Method 2: $y=e^{\cos x\ln \sin x}$

$\Rightarrow \frac{dy}{dx}=e^{\cos x\ln \sin x}\left(\frac{\cos x\cdot \cos x}{\sin x}+(-\sin x)\ln \sin x\right)$

$\Rightarrow y\left(\frac{{\cos }^{2}x}{\sin x}-\sin x\cdot \ln \sin x\right)$

Example 4: Differentiate $\sin x\text{ w.r.t. }{x}^3$

Solution:

$\frac{d(\sin x)}{d\left(x^3\right)}=\frac{\cos x}{3x^2}$

Example 5: If $x=\frac{1-t^2}{1+t^2}\text{ and }y=\frac{2at}{1+t^2}\text{, then }\frac{dy}{dx}$=

Solution:

$x=\frac{1-t^2}{1+t^2}\text{ and }y=\frac{2at}{1+t^2}$

Differentiating with respect to “t”, we get

$\frac{dx}{dt}=\frac{\left(1+t^2\right)(0-2t)-\left(1-t^2\right)(0+2t)}{{\left(1+t^2\right)}^2}=-\frac{4t}{{\left(1+t^2\right)}^2}$

and

$\frac{dy}{dt}=\frac{\left(1+t^2\right)2a-2at(2t)}{{\left(1+t^2\right)}^2}=\frac{2a\left(1-t^2\right)}{{\left(1+t^2\right)}^2}$

$\Rightarrow \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{a\left(1-t^2\right)}{-2t}$

$\Rightarrow \frac{dy}{dx}=\frac{a\left(t^2-1\right)}{2t}$.

Example 6: If $x\sqrt{1+y}+y\sqrt{1+x}=0$, then prove that $\frac{dy}{dx}=-\frac{1}{(1+x{)}^2}$ (explicit form)

Solution:

$(x\sqrt{1+y}{)}^2=(y\sqrt{1+x}{)}^2\Rightarrow xy(x-y)+\left(x^2-y^2\right)=0\text{ or }(x-y)(xy+x+y)=0$

$\therefore y=-\frac{x}{1+x}=\frac{1}{1+x}-1$ (as y = x does not satisfy)

hence $y^{\prime }=-\frac{1}{(1+x{)}^2}$

Example 7: Find the derivative of tan (2x + 3) 

Solution:

Let f(x) = tan (2x + 3), u(x) = 2x + 3 and v(t) = tan t. 

Then (v o u) = v(u(x)) = v (2x + 3) = tan (2x + 3) = f(x)

As a result, f is a composite of two functions. Put t = u(x) = 2x + 3.

Then $\frac{\mathrm{dv}}{\mathrm{dt}}=sec2t$ and $\frac{\mathrm{dt}}{\mathrm{dx}}=2$ exist. Hence, by chain rule $\frac{\mathrm{df}}{\mathrm{dx}}=\frac{\mathrm{dv}}{\mathrm{dt}}\cdot \frac{\mathrm{dt}}{\mathrm{dx}}=2{\sec }^2(2x+3)$

10.0Practice Problems on Differentiation

1. Find the derivative of the function $f(x)=(x^2+3x)\cdot sin(x)$
2. Find the derivative of the function $f(x)=\frac{x^2+2x+1}{x^2-1}$
3. .Find the derivative of the function $f(x)=e^{2x}\cdot \cos (x)$
4. Find the derivative of the function $f(x)=\ln \left(3x^2+4x\right)$
5. Find the derivative of the function $f(x)=\sqrt{x}\cdot \cos (x).$
6. Find the derivative of the function $f(x)=x^3\cdot e^x.$