RD Sharma Class 11 Solutions Chapter 30 - Derivatives (Ex 30.3) Exercise 30.3 - Free PDF
FAQs on RD Sharma Class 11 Solutions Chapter 30 - Derivatives (Ex 30.3) Exercise 30.3
1. What is the fundamental rule used to find the derivative of a product of two functions, such as x sin(x), in Exercise 30.3?
To find the derivative of a product of two functions, like f(x) = u(x)v(x), you must use the product rule. The formula is d/dx(uv) = u(dv/dx) + v(du/dx). For the function x sin(x), you would identify u = x and v = sin(x), find their individual derivatives (du/dx = 1, dv/dx = cos(x)), and substitute them into the formula to get the final answer: 1⋅sin(x) + x⋅cos(x).
2. How do you correctly apply the quotient rule for differentiating functions in the form of f(x)/g(x), as seen in RD Sharma Ex 30.3?
The quotient rule is essential for differentiating functions that are fractions. The correct step-by-step method is:
1. Identify the numerator function (u) and the denominator function (v).
2. Calculate the derivative of the numerator (u') and the derivative of the denominator (v').
3. Apply the formula: d/dx (u/v) = (u'v - uv') / v².
A common mistake is reversing the terms in the numerator, so always ensure it is u'v minus uv'.
3. What are the standard derivatives of the six trigonometric functions required for solving problems in Chapter 30?
To solve problems in this chapter efficiently, you should memorize the following standard derivatives as per the CBSE syllabus:
d/dx (sin x) = cos x
d/dx (cos x) = -sin x
d/dx (tan x) = sec² x
d/dx (cot x) = -cosec² x
d/dx (sec x) = sec x tan x
d/dx (cosec x) = -cosec x cot x
4. Why does the derivative of cos(x) have a negative sign (-sin x), while the derivative of sin(x) is positive (cos x)?
The negative sign in the derivative of cos(x) reflects its rate of change. On the unit circle, as the angle x increases from 0 to π/2, the value of sin(x) (the y-coordinate) increases, so its slope (derivative) is positive. In contrast, the value of cos(x) (the x-coordinate) decreases over this same interval, meaning its rate of change, or slope, must be negative. This fundamental difference in their behaviour is why their derivatives have opposite signs.
5. How are the derivatives of sec(x), cosec(x), and cot(x) derived using the derivatives of sin(x) and cos(x)?
The derivatives of sec(x), cosec(x), and cot(x) can be found using the quotient rule and their definitions in terms of sin(x) and cos(x).
For example, to find the derivative of sec(x):
1. Write sec(x) as 1/cos(x).
2. Apply the quotient rule with u=1 and v=cos(x).
3. This results in (0⋅cos(x) - 1⋅(-sin(x))) / cos²(x) = sin(x)/cos²(x).
4. Simplifying this gives (1/cos(x))⋅(sin(x)/cos(x)) = sec(x)tan(x). A similar process is used for cosec(x) and cot(x).
6. When solving problems from Exercise 30.3, how does the chain rule interact with trigonometric functions?
The chain rule is applied when you have a composite function, i.e., a function inside another function. For trigonometric functions, this looks like sin(ax+b) or tan(x²). The process is to first differentiate the outer trigonometric function, keeping the inner function unchanged, and then multiply it by the derivative of the inner function. For example, the derivative of cos(3x) is [-sin(3x)] multiplied by the derivative of 3x, giving the final answer -3sin(3x).
7. What is a common conceptual error students make when differentiating a term like 'sin(a)' where 'a' is a constant?
A very common error is to mistakenly differentiate a constant trigonometric term. For example, in the expression y = x² + sin(a), students might incorrectly write the derivative as 2x + cos(a). However, since 'a' is a constant, sin(a) is also a constant. The derivative of any constant is zero. Therefore, the correct derivative of y = x² + sin(a) is simply 2x. Always check if the variable inside the trigonometric function is the variable of differentiation (like 'x') or a constant.
