Differentiation and Higher Order Derivatives Short Notes for GATE DA
Differentiation and Higher Order Derivatives short notes for GATE DA: 1 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practic
differentiation and higher order derivatives short notes
Quick Revision Cheat Sheet
Essential Derivatives Checklist
Power
dxdxn=nxn−1
Exp
dxdex=ex ; dxdax=axlna
Log
dxdlnx=x1
Trig
sin→cos ; cos→−sin ; tan→sec2
Inv Trig
sin−1→1−x21 ; tan−1→1+x21
Special
Sigmoid: f′=f(1−f)
Hyperbolic
sinh→cosh ; cosh→sinh
Final Tip: Always check for chain rule application after identifying the standard form.
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Question 1
Level 1: Warm-up
How many of the following functions have a derivative that is a rational function (a ratio of two polynomials)?
1. f(x)=x−2
2. g(x)=x1/2
3. h(x)=x−3/2
4. k(x)=x3
Question 2
Level 1: Warm-up
Let f(x)=x3/2. Which of the following values is IMPOSSIBLE for its derivative f′(x)?
Question 3
Level 1: Warm-up
What is the minimum value of f′(x) where f(x)=x3−3x on the interval [−1,1]?
Question 4
Level 1: Warm-up
Match each function in List I with its correct derivative in List II.
List I
P. f(x)=2x
Q. g(x)=e−x
R. h(x)=xe
S. k(x)=ex
List II
1. ex
2. −e−x
3. 2xln(2)
4. exe−1
Select the correct matching:
Question 5
Level 1: Warm-up
Let f(x)=x−3. Which of the following values is IMPOSSIBLE for its derivative f′(x)?
Question 6
Level 1: Warm-up
Let f1(x)=sin(x), f2(x)=cos(x), f3(x)=tan(x), and f4(x)=cot(x).
Evaluate their derivatives at x=4π and rank the values from MOST NEGATIVE to MOST POSITIVE:
P. f1′(4π)
Q. f2′(4π)
R. f3′(4π)
S. f4′(4π)
Question 7
Level 1: Warm-up
Consider the following two statements regarding the derivative of f(x)=sec(x):
Assertion (A): The derivative of the function is f′(x)=sec(x)tan(x).
Reason (R): The secant function is a co-function, and its derivative follows the same negative sign pattern as other co-functions like cosine and cotangent.
Which of the following is correct?
Question 8
Level 1: Warm-up
How many of the following functions have a derivative that is ALWAYS positive for all x in their domain?
What is the minimum value of f′(x) where f(x)=sin(x)−x on the interval [0,π]?
Question 10
Level 1: Warm-up
Consider the following two statements regarding the derivative of f(x)=x2 at x=3:
Assertion (A): The derivative can be evaluated using the limit limh→0h(3+h)2−32.
Reason (R): The limit definition of a derivative is a fallback method used only when the standard power rule fails.
Which of the following is correct?
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Differentiation and Higher Order Derivatives Short Notes for GATE DA
Differentiation and Higher Order Derivatives short notes for GATE DA: 1 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.
Quick Revision Cheat Sheet
Essential Derivatives Checklist
Power
dxdxn=nxn−1
Exp
dxdex=ex ; dxdax=axlna
Log
dxdlnx=x1
Trig
sin→cos ; cos→−sin ; tan→sec2
Inv Trig
sin−1→1−x21 ; tan−1→1+x21
Special
Sigmoid: f′=f(1−f)
Hyperbolic
sinh→cosh ; cosh→sinh
Final Tip: Always check for chain rule application after identifying the standard form.
Differentiation and Higher Order Derivatives: Solved Questions with Step-by-Step Explanations (10 Problems)
Question 1 · Calculus and OptimizationNAT
How many of the following functions have a derivative that is a rational function (a ratio of two polynomials)?
1. f(x)=x−2
2. g(x)=x1/2
3. h(x)=x−3/2
4. k(x)=x3
Correct Answer:
2.00
Step-by-Step Solution
Key idea: A rational function must only have integer powers of x in the numerator and denominator.
Step 1: Derivative of f(x)=x−2
f′(x)=−2x−3=x3−2. This is a ratio of polynomials. (Yes)
Step 2: Derivative of g(x)=x1/2
g′(x)=21x−1/2=2x1. The denominator contains a square root, so it is not a polynomial. (No)
Step 3: Derivative of h(x)=x−3/2
h′(x)=−23x−5/2=2x2x−3. Contains a square root. (No)
Step 4: Derivative of k(x)=x3
k′(x)=3x2=13x2. This is a polynomial, which is a rational function. (Yes)
Count: Functions 1 and 4 have rational derivatives.
Answer: 2.00
Question 2 · Calculus and OptimizationMCQ
Let f(x)=x3/2. Which of the following values is IMPOSSIBLE for its derivative f′(x)?
A.
1.50
B.
3.00
C.
-1.50
D.
0.00
Correct Answer:
C
Step-by-Step Solution
Key idea: Find the derivative and determine its range based on the domain of the original function.
Step 1: Compute the derivative
f(x)=x3/2
f′(x)=23x1/2=23x
Step 2: Analyze the domain and range
The domain of f(x)=x3/2 is x≥0 (for real-valued functions).
For x≥0, the square root x≥0.
Therefore, f′(x)=23x≥0.
Step 3: Check the options
The derivative must be greater than or equal to 0.
1.50 is possible (at x=1)
3.00 is possible (at x=4)
-1.50 is IMPOSSIBLE (negative)
0.00 is possible (at x=0)
Answer: -1.50
Question 3 · Calculus and OptimizationNAT
What is the minimum value of f′(x) where f(x)=x3−3x on the interval [−1,1]?
Correct Answer:
-3.00
Step-by-Step Solution
Key idea: Find the derivative using the power rule and determine its minimum on the given closed interval.
Step 1: Compute the derivative
f(x)=x3−3x
f′(x)=3x2−3
Step 2: Analyze f′(x) on [−1,1]
The function f′(x)=3x2−3 is a parabola opening upwards.
Its vertex is at x=0.
Since x=0 is within the interval [−1,1], the minimum occurs at the vertex.
Step 3: Evaluate at the minimum
f′(0)=3(0)2−3=−3
(Check endpoints just in case: f′(−1)=3(1)−3=0, f′(1)=3(1)−3=0. The minimum is indeed -3.)
Answer: -3.00
Question 4 · Calculus and OptimizationMCQ
Match each function in List I with its correct derivative in List II.
List I
P. f(x)=2x
Q. g(x)=e−x
R. h(x)=xe
S. k(x)=ex
List II
1. ex
2. −e−x
3. 2xln(2)
4. exe−1
Select the correct matching:
A.
P→3, Q→2, R→4, S→1
B.
P→1, Q→2, R→4, S→3
C.
P→3, Q→1, R→4, S→2
D.
P→4, Q→2, R→3, S→1
Correct Answer:
A
Step-by-Step Solution
Key idea: Identify whether the variable is in the base or the exponent to apply the correct rule.
Step 1: f(x)=2x
Variable x is in the exponent. Use the exponential rule: dxd(ax)=axln(a).
f′(x)=2xln(2). Matches 3.
Step 2: g(x)=e−x
Use the chain rule with the exponential rule: dxd(eu)=eu⋅u′.
g′(x)=e−x⋅(−1)=−e−x. Matches 2.
Step 3: h(x)=xe
Variable x is in the base. Use the power rule: dxd(xn)=nxn−1.
h′(x)=exe−1. Matches 4.
Step 4: k(x)=ex
The natural exponential is its own derivative.
k′(x)=ex. Matches 1.
Answer: P→3, Q→2, R→4, S→1
Question 5 · Calculus and OptimizationMCQ
Let f(x)=x−3. Which of the following values is IMPOSSIBLE for its derivative f′(x)?
A.
-3.00
B.
-0.75
C.
0.00
D.
-12.00
Correct Answer:
C
Step-by-Step Solution
Key idea: Find the derivative and determine its range based on the domain of the function.
Step 1: Compute the derivative
f(x)=x−3
f′(x)=−3x−4=x4−3
Step 2: Analyze the domain and range
The domain of f(x) is all real x=0.
For any x=0, x4>0.
Therefore, the numerator is −3 (negative) and the denominator is positive.
This means f′(x)<0 for all x in the domain.
Step 3: Check the options
The derivative must be strictly negative.
-3.00 is possible (at x=1)
-0.75 is possible (at x=2)
0.00 is IMPOSSIBLE (it can never be zero)
-12.00 is possible (at x=1/2)
Answer: 0.00
Question 6 · Calculus and OptimizationMCQ
Let f1(x)=sin(x), f2(x)=cos(x), f3(x)=tan(x), and f4(x)=cot(x).
Evaluate their derivatives at x=4π and rank the values from MOST NEGATIVE to MOST POSITIVE:
P. f1′(4π)
Q. f2′(4π)
R. f3′(4π)
S. f4′(4π)
A.
S, Q, P, R
B.
S, P, Q, R
C.
Q, S, P, R
D.
P, Q, R, S
Correct Answer:
A
Step-by-Step Solution
Key idea: Compute each derivative and evaluate at x=π/4 using exact trigonometric values.
Step 1: f1(x)=sin(x)
f1′(x)=cos(x)
f1′(π/4)=cos(π/4)=21≈0.707
Step 2: f2(x)=cos(x)
f2′(x)=−sin(x)
f2′(π/4)=−sin(π/4)=−21≈−0.707
Step 3: f3(x)=tan(x)
f3′(x)=sec2(x)
f3′(π/4)=sec2(π/4)=(2)2=2
Step 4: f4(x)=cot(x)
f4′(x)=−csc2(x)
f4′(π/4)=−csc2(π/4)=−(2)2=−2
Step 5: Rank from most negative to most positive
S: −2
Q: −0.707
P: 0.707
R: 2
Order: S, Q, P, R
Answer: S, Q, P, R
Question 7 · Calculus and OptimizationMCQ
Consider the following two statements regarding the derivative of f(x)=sec(x):
Assertion (A): The derivative of the function is f′(x)=sec(x)tan(x).
Reason (R): The secant function is a co-function, and its derivative follows the same negative sign pattern as other co-functions like cosine and cotangent.
Which of the following is correct?
A.
Both A and R are true, and R is the correct explanation of A
B.
Both A and R are true, but R is NOT the correct explanation of A
C.
A is false but R is true
D.
A is true but R is false
Correct Answer:
D
Step-by-Step Solution
Key idea: Verify the standard derivative formula and check the sign pattern for co-functions.
Step 1: Analyze Assertion (A)
The standard derivative of sec(x) is indeed sec(x)tan(x).
Thus, A is TRUE.
Step 2: Analyze Reason (R)
The reason claims that secant follows the "negative sign pattern" of co-functions.
While it is true that cos(x) and cot(x) have negative derivatives, sec(x) is an exception. Its derivative sec(x)tan(x) is positive in the first quadrant.
Thus, R is FALSE.
Step 3: Conclusion
A is true, but R is false.
Answer: A is true but R is false.
Question 8 · Calculus and OptimizationNAT
How many of the following functions have a derivative that is ALWAYS positive for all x in their domain?
Key idea: Compute the derivative of each function and check if it is strictly positive over its entire domain.
Step 1: f(x)=tan(x)
f′(x)=sec2(x). Since sec2(x)>0 for all x in the domain of tan(x), this is ALWAYS positive. (Yes)
Step 2: g(x)=cot(x)
g′(x)=−csc2(x). This is always negative. (No)
Step 3: h(x)=sec(x)
h′(x)=sec(x)tan(x). This changes sign depending on the quadrant (e.g., positive in Q1, negative in Q2). (No)
Step 4: k(x)=csc(x)
k′(x)=−csc(x)cot(x). This also changes sign depending on the quadrant. (No)
Step 5: m(x)=tan(x)+x
m′(x)=sec2(x)+1. Since sec2(x)>0, the sum is always >1, so it is ALWAYS positive. (Yes)
Count: Functions 1 and 5 have always-positive derivatives.
Answer: 2.00
Question 9 · Calculus and OptimizationNAT
What is the minimum value of f′(x) where f(x)=sin(x)−x on the interval [0,π]?
Correct Answer:
-2.00
Step-by-Step Solution
Key idea: Find the derivative using standard rules and determine its minimum on the given closed interval.
Step 1: Compute the derivative
f(x)=sin(x)−x
f′(x)=cos(x)−1
Step 2: Analyze f′(x) on [0,π]
We need to find the minimum of g(x)=cos(x)−1 on [0,π].
On the interval [0,π], the function cos(x) decreases from 1 to -1.
Therefore, the minimum value of cos(x) on this interval is -1 (which occurs at x=π).
Step 3: Find the minimum of f′(x)
Since the minimum of cos(x) is -1, the minimum of cos(x)−1 is:
−1−1=−2
Answer: -2.00
Question 10 · Calculus and OptimizationMCQ
Consider the following two statements regarding the derivative of f(x)=x2 at x=3:
Assertion (A): The derivative can be evaluated using the limit limh→0h(3+h)2−32.
Reason (R): The limit definition of a derivative is a fallback method used only when the standard power rule fails.
Which of the following is correct?
A.
Both A and R are true, and R is the correct explanation of A
B.
Both A and R are true, but R is NOT the correct explanation of A
C.
A is false but R is true
D.
A is true but R is false
Correct Answer:
D
Step-by-Step Solution
Key idea: The limit definition is the fundamental definition of a derivative, not a fallback.
Step 1: Analyze Assertion (A)
The limit limh→0hf(x+h)−f(x) is the first-principles definition of f′(x).
For f(x)=x2 at x=3, this limit correctly evaluates to 2(3)=6. Thus, A is TRUE.
Step 2: Analyze Reason (R)
The reason claims the limit definition is only used when the power rule fails.
This is FALSE. The limit definition is universally applicable and is the foundation from which the power rule is derived. We use the power rule for speed, not because the limit definition is invalid.