Rachit Kumar SaxenaManager-Editorial
The differentiation of a capacity f(x) is spoken to as f'(x). In the event that f(x) = y, at that point f'(x) = dy/dx, which implies y is separated as for x. Before we begin tackling a few inquiries dependent on differentiation, let us see the overall differentiation equations utilized here.
Function f(x) = y | Differentiation of function f’(x) = dy/dx |
---|---|
xn | nxn-1 |
ex | ex |
ln(x) | 1/x |
sin x | cos x |
cos x | -sin x |
tan x | sec2x |
K (constant) | 0 |
What are Differential Questions?
1. Differentiate y = (tan x - 1)/ sec x
y = (tan x - 1)/ sec x
u = tan x - 1 ===> u' = sec2 x - 0
v = sec x ===> v' = sec x tan x
2.Differentiate y = sin x / x2
y = sin x / x2
u = sin x ==> u' = cos x
u = x2 ==> v' = 2 x
dy/dx = [x2 (cos x) - sin x(2x)]/(x2)2
dy/dx = [x2 (cos x) - 2x sin x]/x4
dy/dx = x [x cos x - 2 sin x]/x4
dy/dx = [x cos x - 2 sin x]/x3
3. Differentiate y = tan θ (sin θ + cos θ)
y = tan θ (sin θ + cos θ)
u = tan θ ==> u' = sec2 θ
v = sin θ + cos θ ==> v' = cos θ - sin θ
dy/dx = tan θ (cos θ - sin θ) + (sin θ + cos θ) (sec2θ)
= tan θ cos θ - tan θ sin θ + sin θ sec2θ + cos θ sec2θ
4. Differentiate y = cosec x ⋅ cot x
y = cosec x ⋅ cot x
u = cosec x ==> u' = -cosec x cot x
v = cot x ==> v' = -cosec2 x
dy/dx = cosec x (-cosec2 x) + cot x(-cosec x cot x)
= - cosec3 x - cosec x cot2 x
= - (1/sin3 x) - (1/sin x) (cos2 x/sin2 x)
= - (1/sin3 x) - (cos2 x/sin3 x)
dy/dx = - (1 + cos2x) /sin3 x
5. Differentiate 10x2 with respect to x.
y = 10x2
y’ = d(10x2)/dx
y’ = 2.10.x = 20x
Therefore, d(10x2)/dx = 20 x
6. Compute the derivative of f(x) = sin2x.
f(x) = sin2x = sin x sin x
= d(sin x)/dx. sin x + sin x.d(sin x)/dx
= cos x. sin x + sin x cos x
= 2 sin x cos x
7. Differentiate tan2x.
Say, y = tan2x
dy/dx = d(tan2x)/dx
= 2tan2-1x. d(tan x)/dx
= 2tan x sec2x
d[tan2x]/dx = 2tan x sec2x
Weightage of Differential Questions
In Class 11: The topic has been discussed with the basic explanations and the ways to solve the particular question. It has a weightage of 30 Marks.
Illustrated Examples on Differential Questions
1. Differentiate sin(3x+5)
Solution.
Say, y = sin (3x+5)
dy/dx = d[sin(3x+5)]/dx
= cos (3x+5) d(3x+5)/dx [ By chain rule]
= cos (3x+5) [3]
y’ = 3 cos (3x+5)
d[sin(3x+5)]/dx = 3 cos (3x+5)
2.Differentiate 20x-4 + 9.
Solution.
y = 20x-4 + 9
y’ = d(20x-4 + 9)/dx
y’ = d(20x-4)/dx + d(9)/dx
y’ = -4.20.x-4-1+0
y’ = -80x-5
Therefore, d(20x-4 + 9)/dx = -80x-5
3. Differentiate x5 with respect to x.
Solution.
Given, y = x5
On differentiating w.r.t we get;
dy/dx = d(x5)/dx
y’ = 5x5-1 = 5x4
Therefore, d(x5)/dx = 5x4=20
FAQs on Differential Questions
Q: What do you mean by differentiation?
Q: Differentiate ln(10).
Q: How can we differentiate log?
Q: What is an equation to solve?
Q: What are the types of equations?
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