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Derivatives are a fundamental concept in calculus that measures how a function changes as its input changes. They represent the rate of change or slope of the function at any given point. Understanding derivatives is crucial for solving problems related to motion, optimization, and modeling real-world phenomena. In Chapter 30 of RD Sharma's Class 11 textbook, students delve into the basics of derivatives including the rules and applications which are essential for mastering calculus.
The Derivatives provide a way to quantify the rate at which one quantity changes with respect to another. For a function f(x) the derivative 𝑓(x) represents the instantaneous rate of the change of f(x) concerning x. This concept is foundational in mathematics and is widely used in physics, engineering, economics, and various other fields. The Derivatives help in understanding the behavior of functions finding tangents to curves and solving the problems related to rates of change.
Solution:
Given that f(x) = xsinx
By using the formula
We get
=
=
Using the formula
sinc - sind = 2cos((c + d)/2)sin((c - d)/2)
We get
=
As we know that
So,
= 2x × cosx × 1/2 + sinx
= x × cosx + sinx
= sinx + xcosx
Solution:
Given that f(x) = xcosx
By using the formula
We get
=
=
=
=
=
= -xsinx + cosx
Solution:
Given that f(x) = sin(2x - 3)
By using the formula
We get
=
=
Using the formula
sinC - sinD = 2cos{C+D}/2sin{C-D}/2
=
As we know that, \lim_{θ\to 0}\frac{sinθ}{θ}=1 so,
= 2cos(2x - 3)
Solution:
Given that f(x) = √sin2x
By using the formula
We get
=
On multiplying numerator and denominator by
we get
=
=
=
=
=
Solution:
Given that f{x} = sinx/x
By using the formula
We get
=
=
=
=
=
h ⇢ 0 ⇒ h/2 ⇢ 0 and
=
=
Solution:
Given that f(x) = cosx/x
By using the formula
We get
=
=
=
=
=
=
=
Solution:
Given that f(x) = x2sinx
By using the formula
We get
=
=
=
=
= 0 + [2xsinx + x2cosx]
= 2xsinx + x2cosx
Solution:
Given that f(x) =
By using the formula
We get
=
=
=
=
=
Solution:
Given that f(x) = sinx + cosx
By using the formula
We get
=
=
=
=
=
=
= cosx - sinx
Solution:
Given that f(x) = tan2x
By using the formula
We get
=
=
=
=
=
=
=
= 2tanx sec2x
Solution:
Given that f(x) = tan(2x+1)
By using the formula
We get
=
=
=
Multiplying both, numerator and denominator by 2.
=
=
= 2sec2(2x+1)
Solution:
Given that f(x) = tan2x
By using the formula
We get
=
=
=
=
=
= 2sec22x
Solution:
Given that f(x) = √tanx
By using the formula
We get
=
On multiplying numerator and denominator by
We get
=
=
=
=
=
Solution:
Given that f(x) =
By using the formula
We get
=
=
=
=
=
=
Solution:
Given that f(x) = cos√x
By using the formula
We get
=
=
=
Multiplying numerator and denominator by
=
=
=
=
Solution:
Given that f(x) = tan√x
By using the formula
We get
=
=
=
=
=
=
=
=
Solution:
Given that f(x) = tanx2
By using the formula
We get
=
=
=
=
=
=
=
=
= 2xsec2x2
Solution:
Given that f(x) = -x
By using the formula
We get
=
=
= -1
Solution:
Given that f(x) = (-x)-1
By using the formula
We get
=
=
=
= 1/x2
Solution:
Given that f(x) = sin(x+1)
By using the formula
We get
=
=
=
=
=
= cos(x+1)
Solution:
We have, f(x) = cos(x - π/8)
By using the formula
We get
=
=
=
=
=
=
= -sin(x + π/8)
Read More:
Exercise 30.2 Set 2 covers differentiation of various functions, including polynomials, trigonometric functions, exponential functions, and logarithmic functions. Students learn to apply differentiation rules and formulas to find derivatives. Understanding derivatives is crucial for calculus and its applications. Differentiation helps analyze functions and model real-world phenomena. Practice questions reinforce learning and application. Derivatives measure rates of change, essential for optimization and physics.