• Introduction

Let {y = f(x)} be a real-valued function of {x}. Consider the following limit :

{\lim \limits_{h \to 0} \frac {f(x+h)-f(x)}{(x+h) - (x)} = \frac {f(x+h)-f(x)}{h}}

It asks the following question : If {h} is the change in value of {x}, what will be the fractional change in {f(x)}, as the change in {h \to 0}?

If this limit exists, it is known as the derivative of {f(x)} w.r.t. {x} and is denoted by

{\frac {dy}{dx} \ or \ f'(x)}

It is also known as the differential coefficient.

The process of obtaining derivatives of function using the evaluation of above limit is known as derivative using first principle. However, while solving problems, we will use the formulas of derivatives of standard functions directly.

  • Differentiable Function

When the right hand derivative

{\lim \limits_{h \to 0^{+}} \frac {f(a+h)-f(a)}{h}}

and the left hand derivative

{\lim \limits_{h \to 0^{-}} \frac {f(a+h)-f(a)}{h}}

exist and are equal, then the function is said to be differentiable at {x = a}.

  • Theorems

I) If a function is differentiable at a point, then it is continuous at that point.

II) If a function is continuous at a point, then it may not be always differentiable at that point.

In other words, differentiability is a sufficient condition for continuity. However, continuity is a necessary condition for differentiability.

  • Chain Rule of Differentiation

We know that a composite function is a function of a function of {x}. For example,

{f(x) = sin (x^4)}

{x^4} is a function and {f(x)} is the {sine} of {x^4}.

We use chain rule to get the derivatives of such functions.

If {y = f(u)} is a differentiable function of {u} and {u=g(x)} is a differentiable function of {x}, then,

{\frac {dy}{dx} = \frac {dy}{du} \times \frac {du}{dx}}

  • Derivative of Inverse Function

Let {y=f(x)} be a differentiable function. If the inverse function {x = f^{-1}y} exists, then,

{\frac {dx}{dy} = \frac {1}{ \frac {dy}{dx}}, \ s.t. \ \frac {dy}{dx} \ne 0}

  • Derivatives of Inverse Trigonometric Functions

If {y = sin^{-1}x , \ -1 \le x \le 1, \ \frac {- \pi}{2} \le y \le \frac {\pi}{2}}, then {\frac {dy}{dx} = \frac {1}{\sqrt {1 - x^2}}}

If {y = cos^{-1}x , \ -1 \le x \le 1, \ 0 \le y \le \pi}, then {\frac {dy}{dx} = \frac {-1}{\sqrt {1 - x^2}}}

If {y= tan^{-1}x, \ x \in \mathbb R, \ \frac {- \pi}{2} < y < \frac {\pi}{2}}, then {\frac {dy}{dx} = \frac {1}{1+x^2}}

If {y= cot^{-1}x, \ x \in \mathbb R, \ 0 < y < \pi}, then {\frac {dy}{dx} = \frac {-1}{1+x^2}}

If {y = sec^{-1}x, \ |x| \ge 1, \ 0 \le y \le \pi, \ y \ne \frac {\pi}{2}}, then

{\frac {dy}{dx} = \frac {1}{x \sqrt {x^2-1}}, x > 1,}

{\frac {dy}{dx} = \frac {-1}{x \sqrt {x^2-1}}, x < -1}

If {y = cosec^{-1}x, \ |x| \ge 1, \ , \frac {- \pi}{2} \le y \le \frac {\pi}{2}\ y \ne 0}, then

{\frac {dy}{dx} = \frac {-1}{x \sqrt {x^2-1}}, x > 1,}

{\frac {dy}{dx} = \frac {1}{x \sqrt {x^2-1}}, x < -1}

  • Important Substitutions to Solve Problems on Inverse Trigonometric Functions



  • Logarithmic Differentiation

This technique uses logarithms to get the derivatives. To get derivatives of functions of the form {[f(x)]^{g(x)}}, we use this.

Important: {\frac {d}{dx} log (y) = \frac {1}{y} \frac {dy}{dx}}

Properties of logarithms :

I) {log (a \times b) = log (a) + log (b)}

II) {log (a/b) = log (a) - log (b)}

III) {log (a^m) = m \times log (a)}

  • Implicit Functions and Their Derivatives

An implicit (or indirect) function is the one, which is of the form {f(x,y)= 0}. In other words, one cannot write {y} as a function, which contains only terms of {x} and not {y}. For example,

{x^m \cdot y^n = (x+y)^{m+n}}

The derivative is to be obtained by using chain rule and sometimes using logarithmic differentiation.

Useful : For {x^m \cdot y^n = (x+y)^{m+n}}, {\frac {dy}{dx} = \frac {y}{x}}.

  • Derivatives of Parametric Functions

If {f(t)} and {g(t)} are differentiable functions of parameter {t}, then,

{\frac {dy}{dx} = \frac {\frac {dy}{dt}}{\frac {dx}{dt}} , \frac {dx}{dt} \ne 0}

  • Derivatives of Higher Order

Order : Number of times {y} is differentiated w.r.t. {x}

First order : {\frac {dy}{dx} = y_1},

Second order : {\frac {d}{dx} \frac {dy}{dx} = \frac {d^2y}{dx^2} = y_2},

Third order : {\frac {d}{dx} \frac {d^2y}{dx^2} = \frac {d^3y}{dx^3} = y_3},

and so on.

Higher order derivatives come into picture, when many natural phenomena like heat transfer are modeled.


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s