TMUA: Differentiating powers of x
Differentiating powers of \(x\).
- Section 1 Part 1 · MM6 Differentiation · spec MM6.1-MM6.2
- Papers 1 and 2
- 7 practice questions
- No calculator
What the specification covers
- Derivative as gradient and rate of change; second derivatives
- Rational powers after simplifying (no first principles)
Key ideas
- \(\dfrac{d}{dx}x^n=nx^{n-1}\) for any rational \(n\).
- Rewrite first: \(\dfrac{x^2+4}{\sqrt x}=x^{3/2}+4x^{-1/2}\).
- \(f'(x)\) is the gradient of the tangent and the rate of change; \(f''(x)\) is the rate of change of the gradient.
Common mistakes
- You cannot differentiate a quotient term by term: split it first.
- \(\tfrac{d}{dx}x^{-1/2}=-\tfrac12x^{-3/2}\): subtract 1 from the power.
Exam tip
Always convert roots and fractions to powers before differentiating.
Worked example
Worked example
Given \(f(x)=\dfrac{x^2+4}{\sqrt{x}}\) for \(x>0\), what is \(f'(4)\)?
- \(\frac{7}{4}\)
- \(\frac{11}{4}\)
- \(\frac{13}{4}\)
- \(\frac{5}{2}\)
- \(3\)
Answer: B: \(\frac{11}{4}\)
\(f(x)=x^{3/2}+4x^{-1/2}\), so \(f'(x)=\tfrac32x^{1/2}-2x^{-3/2}\). \(f'(4)=3-\tfrac{2}{8}=\tfrac{11}{4}\).
Practice questions
Try each question before opening the solution. No calculator: the TMUA does not allow one. Every answer here was re-checked independently by computer before it was published.
Question 1
Given \(f(x)=\dfrac{x^2+1}{\sqrt{x}}\) for \(x>0\), what is \(f'(9)\)?
- \(\frac{121}{27}\)
- \(\frac{9}{2}\)
- \(\frac{122}{27}\)
- \(3\)
- \(\frac{323}{54}\)
Show the answer and solution
Answer: A: \(\frac{121}{27}\)
\(f(x)=x^{3/2}+x^{-1/2}\), so \(f'(x)=\tfrac32x^{1/2}-\tfrac{1}{2}x^{-3/2}\). At \(x=9\): \(\tfrac32\cdot3-\tfrac{1}{2\cdot27}=\frac{121}{27}\).
Question 2
Given \(f(x)=\dfrac{x^2+8}{\sqrt{x}}\) for \(x>0\), what is \(f'(16)\)?
- \(\frac{127}{16}\)
- \(4\)
- \(6\)
- \(\frac{95}{16}\)
- \(\frac{97}{16}\)
Show the answer and solution
Answer: D: \(\frac{95}{16}\)
\(f(x)=x^{3/2}+8x^{-1/2}\), so \(f'(x)=\tfrac32x^{1/2}-\tfrac{8}{2}x^{-3/2}\). At \(x=16\): \(\tfrac32\cdot4-\tfrac{8}{2\cdot64}=\frac{95}{16}\).
Question 3
Given \(f(x)=\dfrac{x^2+16}{\sqrt{x}}\) for \(x>0\), what is \(f'(16)\)?
- \(4\)
- \(\frac{63}{8}\)
- \(\frac{49}{8}\)
- \(6\)
- \(\frac{47}{8}\)
Show the answer and solution
Answer: E: \(\frac{47}{8}\)
\(f(x)=x^{3/2}+16x^{-1/2}\), so \(f'(x)=\tfrac32x^{1/2}-\tfrac{16}{2}x^{-3/2}\). At \(x=16\): \(\tfrac32\cdot4-\tfrac{16}{2\cdot64}=\frac{47}{8}\).
Question 4
Given \(f(x)=\dfrac{x^2+8}{\sqrt{x}}\) for \(x>0\), what is \(f'(9)\)?
- \(\frac{9}{2}\)
- \(3\)
- \(\frac{158}{27}\)
- \(\frac{251}{54}\)
- \(\frac{235}{54}\)
Show the answer and solution
Answer: E: \(\frac{235}{54}\)
\(f(x)=x^{3/2}+8x^{-1/2}\), so \(f'(x)=\tfrac32x^{1/2}-\tfrac{8}{2}x^{-3/2}\). At \(x=9\): \(\tfrac32\cdot3-\tfrac{8}{2\cdot27}=\frac{235}{54}\).
Question 5
Given \(f(x)=\dfrac{x^2+16}{\sqrt{x}}\) for \(x>0\), what is \(f'(9)\)?
- \(\frac{9}{2}\)
- \(\frac{154}{27}\)
- \(3\)
- \(\frac{259}{54}\)
- \(\frac{227}{54}\)
Show the answer and solution
Answer: E: \(\frac{227}{54}\)
\(f(x)=x^{3/2}+16x^{-1/2}\), so \(f'(x)=\tfrac32x^{1/2}-\tfrac{16}{2}x^{-3/2}\). At \(x=9\): \(\tfrac32\cdot3-\tfrac{16}{2\cdot27}=\frac{227}{54}\).
Question 6
Given \(f(x)=\dfrac{x^2+9}{\sqrt{x}}\) for \(x>0\), what is \(f'(4)\)?
- \(\frac{57}{16}\)
- \(3\)
- \(\frac{39}{16}\)
- \(\frac{55}{16}\)
- \(2\)
Show the answer and solution
Answer: C: \(\frac{39}{16}\)
\(f(x)=x^{3/2}+9x^{-1/2}\), so \(f'(x)=\tfrac32x^{1/2}-\tfrac{9}{2}x^{-3/2}\). At \(x=4\): \(\tfrac32\cdot2-\tfrac{9}{2\cdot8}=\frac{39}{16}\).
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