Test: Meaning of Derivatives - Commerce MCQ

d/dx(1/x² ) = -2 (x ^-2-1) = -2* (x^-3) = -2/x³

f'(100) = lim_(h→0) [f(100+h) - f(100)]/h
= lim_(h→0) [99(100+h) - 99(100)]/h
= lim_(h→0) [9900 + 99h - 9900]/h
= lim_(h→0) 99h/h
= lim_(h→0) 99
= 99

We’ll first need to divide the function out and simplify before we take the derivative. Here is the rewritten function.

The derivative is, 

In calculus, the sum rule in differentiation is a method of finding the derivative of a function that is the sum of two other functions for which derivatives exist. This is a part of the linearity of differentiation.

Let f(x) = a, where a is a fixed real number.
So, f(x+h) = a
So, d/dx(f(x)) = lim_(h→0)f[(x+h)−f(x)]/h
= lim_(h→0) (a−a)/h
= 0
Hence, d/dx(a) = 0 , where a is a fixed real number.

• The process of determining the derivative of a function is known as differentiation. It is clearly visible that the basic concept of derivative of a function is closely intertwined with limits. Therefore, it can be expected that the rules of derivatives are similar to that of limits.
• The following rules are a part of algebra of derivatives:
  Consider f and g to be two real-valued functions such that the differentiation of these functions is defined in a common domain.

This is a formula of question.

This is a formula for finding derivative or differentiation which is represented by
but here at the place of x , c is written So this is equal to f'(c)

f(x) = 3x
f’(x) = lim(h→0) [f(x+h) - f(x)]/h
= f(x+h) = 3(x+h)
f’(x) = = lim(h→0) [3(x+h) - 3(x)]/h
Putting x = ⅔
f’(⅔) = lim(h→0) [3(⅔+h) - 3(⅔)]/h
= lim(h→0) [(6+3h)-6]/h
=lim(h→0) 3h/h
lim(h→0) 3
The derivative of the function f’(⅔) = 3