📝 Derivative of x^n for real powers (15 MCQs)

📖 From Calculus • 4. Topics in Differentiation • 15 questions available

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What is Derivative of x^n for real powers?

Definition:
The power rule for differentiation extends to any real number exponent $n$ , stating that the derivative of $x^n$ is $nx^{n-1}$ . This holds true not just for integers but also for rational and irrational exponents, provided $x$ is in the domain where the function is defined. This generalization is proven using logarithmic differentiation or the definition of the derivative.

Example:
Find the derivative of $f(x) = x^{\sqrt{2}}$ . Applying the power rule, $f'(x) = \sqrt{2}x^{\sqrt{2}-1}$ . Similarly, for $g(x) = x^{-3/4}$ , the derivative is $g'(x) = -\frac{3}{4}x^{-7/4}$ .

Reason:
This universal power rule simplifies the differentiation of root functions (like $\sqrt{x} = x^{1/2}$ ) and reciprocal powers, unifying the treatment of polynomial-like terms and allowing consistent application of calculus rules across all real-valued power functions.

Easy

Medium

Hard

📝 All Derivative of x^n for real powers MCQs

Q1. What is the derivative of $f(x)=x^{p}$ where $p$ is a real constant?

p x^{p-1}

✅

x^{p}

p x^{p}

x^{p-1}

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The power rule states that when a constant exponent $p$ multiplies the variable $x$ , the derivative is obtained by bringing the exponent down as a coefficient and decreasing the exponent by one, giving $p x^{p-1}$ . This directly follows from the limit definition of the derivative.

Q2. If $g(x)=x^{3.5}$ and $g'(a)=0$ for some $a>0$ , what must be true about $a$ ?

a

must be 0

a

must be 1

C.No such

a

exists ✅

a

can be any positive number

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Computing the derivative yields $g'(x)=3.5x^{2.5}$ . Since the factor 3.5 is never zero, the only way the product can be zero is if $x^{2.5}=0$ , which occurs at $x=0$ . Because the premise requires $a>0$ , there is no possible $a$ satisfying the condition.

Q3. Which statement about the growth rates of $f(x)=x^{2}$ and $h(x)=x^{5}$ for large $x$ is correct?

f&#039;(x)

grows faster than

h&#039;(x)

h&#039;(x)

grows faster than

f&#039;(x)

✅

C.Both derivatives grow at the same rate

D.Both derivatives decrease as

x

increases

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The derivatives are $f'(x)=2x$ and $h'(x)=5x^{4}$ . As $x$ becomes large, the term $x^{4}$ dominates, making $5x^{4}$ increase far more rapidly than the linear term $2x$ . Thus the derivative of the higher‑power function outpaces the lower‑power one.

Q4. Let $y = (3x^{2}+2)^{4.2}$ . What is $\dfrac{dy}{dx}$ ?

4.2(3x^{2}+2)^{3.2}\cdot6x

4.2(3x^{2}+2)^{4.2}\cdot6x

25.2x(3x^{2}+2)^{3.2}

✅

25.2x(3x^{2}+2)^{4.2}

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Apply the chain rule: differentiate the outer function $(u)^{4.2}$ to get $4.2u^{3.2}$ and multiply by the derivative of the inner function $u=3x^{2}+2$ , which is $6x$ . Multiplying gives $4.2\cdot6x\,(3x^{2}+2)^{3.2}=25.2x(3x^{2}+2)^{3.2}$ .

Q5. For $f(x)=x^{n}$ with real $n$ , which condition guarantees that $f'(x)$ is negative for all $x<0$ ?

n

is an even integer ✅

n

is an odd integer

n>0

n

is a negative odd integer

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The derivative is $f'(x)=n x^{\,n-1}$ . When $n$ is even, $n-1$ is odd, so $x^{\,n-1}$ is negative for negative $x$ . Multiplying by the positive coefficient $n$ yields a negative product, ensuring $f'(x)<0$ for every $x<0$ .

Q6. Using the definition of derivative, compute $\displaystyle\lim_{h\to0}\frac{(5+h)^{2.3}-5^{2.3}}{h}$ .

2.3\cdot5^{1.3}

B.Approximately

18.63

✅

2.3\cdot5^{2.3}

5^{2.3}

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The limit equals the derivative of $x^{2.3}$ at $x=5$ : $f'(5)=2.3\cdot5^{1.3}$ . Numerically, $5^{1.3}\approx8.10$ ; multiplying by 2.3 gives about $18.63$ . Hence the limit evaluates to roughly $18.63$ .

Q7. Consider $f(x)=x^{x}$ for $x>0$ . Which expression correctly represents $f'(x)$ ?

x^{x}(\ln x+1)

✅

x^{x}\ln x

x^{x}(1-\ln x)

x^{x}(\ln x-1)

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Write $f(x)=e^{x\ln x}$ . Differentiating using the product rule inside the exponent gives $\frac{d}{dx}(x\ln x)=\ln x+1$ . Multiplying by the original function yields $f'(x)=e^{x\ln x}(\ln x+1)=x^{x}(\ln x+1)$ .

Q8. If $f(x)=x^{0.5}$ for $x\ge0$ , what is the sign of $f''(x)$ on $(0,\infty)$ ?

A.Positive

B.Negative ✅

C.Zero

D.Undefined

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: First derivative: $f'(x)=\tfrac12x^{-0.5}$ . Differentiating again gives $f''(x)=-\tfrac14x^{-1.5}$ . Since $x^{-1.5}>0$ for $x>0$ and the coefficient $-\tfrac14$ is negative, the second derivative is negative throughout the interval.

Q9. At $x=2$ , which derivative is larger: that of $x^{3}$ or of $x^{2.5}$ ?

A.Derivative of

x^{3}

is larger ✅

B.Derivative of

x^{2.5}

is larger

C.They are equal

D.Cannot be determined without more information

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Compute each derivative at $x=2$ : $(x^{3})' =3x^{2}=12$ ; $(x^{2.5})' =2.5x^{1.5}\approx7.07$ . Since $12>7.07$ , the derivative of the cubic function exceeds that of the $2.5$ power at the given point.

Q10. Given $x^{2}+y^{3}=7$ , find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .

-\dfrac{2x}{3y^{2}}

✅

-\dfrac{3y^{2}}{2x}

\dfrac{2x}{3y^{2}}

\dfrac{3y^{2}}{2x}

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Differentiate implicitly: $2x+3y^{2}\dfrac{dy}{dx}=0$ . Solving for $\dfrac{dy}{dx}$ yields $\dfrac{dy}{dx}=-\dfrac{2x}{3y^{2}}$ . This expression relates the slope of the curve to the current coordinates.

Q11. For the function $f(x)=x^{p}$ with $0<p<1$ , which statement about its monotonicity on $(0,\infty)$ is true?

A.Increasing and concave up

B.Increasing and concave down ✅

C.Decreasing

D.Constant

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The first derivative $f'(x)=p x^{p-1}$ is positive because $p>0$ and $x^{p-1}>0$ for $x>0$ . The second derivative $f''(x)=p(p-1)x^{p-2}$ is negative since $p-1<0$ . Thus the function rises while bending downward, i.e., it is increasing and concave down.

Q12. If $g(x)=k x^{p}$ where $k>0$ and $p$ is a real constant, how does $g'(x)$ compare to the derivative of $f(x)=x^{p}$ ?

g&#039;(x)=k f&#039;(x)

✅

g&#039;(x)=\dfrac{f&#039;(x)}{k}

g&#039;(x)=k^{p} f&#039;(x)

g&#039;(x)=p k x^{p-1}

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Differentiating $g(x)$ gives $g'(x)=k\cdot p x^{p-1}=k\big(p x^{p-1}\big)=k f'(x)$ . The constant factor $k$ simply scales the original derivative, leaving the functional form unchanged.

Q13. The volume of a sphere is $V=\frac{4}{3}\pi r^{3}$ . If the radius is increasing at $0.2\text{ cm/s}$ when $r=5\text{ cm}$ , what is $\dfrac{dV}{dt}$ at that instant?

4\pi

^3

10\pi

^3

15\pi

^3

20\pi

^3

/s ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Differentiate implicitly: $\dfrac{dV}{dt}=4\pi r^{2}\dfrac{dr}{dt}$ . Substituting $r=5$ cm and $\dfrac{dr}{dt}=0.2$ cm/s gives $4\pi(25)(0.2)=20\pi$ cm $^3$ /s.

Q14. Suppose $h(x)=x^{p}$ with $p\neq0$ . If $h'(c)=0$ for some $c\neq0$ , which conclusion follows?

p

must be zero

c

must be zero

C.Such a

c

cannot exist ✅

h

is a constant function

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The derivative is $h'(x)=p x^{p-1}$ . Setting this equal to zero gives $p x^{p-1}=0$ . Since $p\neq0$ by hypothesis, the only way the product can be zero is if $x^{p-1}=0$ , which occurs at $x=0$ . Hence no non‑zero $c$ can satisfy the condition.

Q15. Find the second derivative of $f(x)=x^{\alpha}$ where $\alpha$ is a real constant, and state for which $\alpha$ the second derivative is positive for all $x>0$ .

f&#039;&#039;(x)=\alpha(\alpha-1)x^{\alpha-2}

; positive when

\alpha<0

\alpha>1

f&#039;&#039;(x)=\alpha(\alpha+1)x^{\alpha-2}

; positive when

\alpha>0

f&#039;&#039;(x)=\alpha(\alpha-1)x^{\alpha-2}

; positive when

0<\alpha<1

f&#039;&#039;(x)=\alpha(\alpha-2)x^{\alpha-2}

; positive when

\alpha>2

✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: Differentiating twice yields $f''(x)=\alpha(\alpha-1)x^{\alpha-2}$ . For $x>0$ , the sign of the second derivative matches the sign of the coefficient $\alpha(\alpha-1)$ . This product is positive when either $\alpha<0$ or $\alpha>1$ , giving the required condition.

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