Differentiation

7 methods · 23 questions

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The power rule applies only to terms $x^{k}$ . Before differentiating, expand any bracket or square, split a single fraction over its numerator terms, and convert every root and reciprocal into a fractional or negative power, so that each term differentiates in one step. This avoids the quotient and product rules entirely on TMUA-style rational expressions.

Trigger: an expression to differentiate that is a product, a square, a quotient, or contains $x$ , $n x$ or $1/ x^{n}$ .

Instances:
(i) split a quotient term by term, e.g. $(x^{3} - 5 x^{2}) / (2 x x) = \frac{1}{2} x^{3/2} - \frac{5}{2} x^{1/2}$ ;
(ii) expand a square such as $(2/ x - 1/ (2 x^{2}))^{2}$ into separate power terms before differentiating once or twice;
(iii) multiply out a product of brackets like $(2 x + a) (x - 2 a)^{2}$ into a polynomial, then differentiate.

Linked questions (3)

TMUA 2016 Paper 2 · Q2

f (x) = \frac{( x ^{2} + 5 ) ( 2 x )}{4 x ^{3}}, x > 0

Which one of the following is equal to

f^{'} (x)

A $8 x^{\frac{9}{4}} + \frac{40}{3} x^{\frac{1}{4}}$
B $\frac{9}{2} x^{\frac{5}{4}} + \frac{5}{2} x^{- \frac{3}{4}}$
C $8 x^{\frac{9}{4}} + \frac{40}{3} x^{- \frac{1}{4}}$
D $\frac{8}{13} x^{\frac{13}{4}} + 8 x^{\frac{5}{4}}$

Worked solution

Answer:B

Simplify

f (x)

f (x) = \frac{2 x ^{3} + 10 x}{x ^{3/4}} = 2 x^{3 - 3/4} + 10 x^{1 - 3/4} = 2 x^{9/4} + 10 x^{1/4} .

Differentiate:

f^{'} (x) = 2 \cdot \frac{9}{4} x^{5/4} + 10 \cdot \frac{1}{4} x^{- 3/4} = \frac{9}{2} x^{5/4} + \frac{5}{2} x^{- 3/4} .

Hence the answer is B.

TMUA 2017 Paper 1 · Q2

The function

f

is given by

f (x) = (\frac{2}{x} - \frac{1}{2 x ^{2}})^{2} (x \neq = 0) .

What is the value of

f^{''} (1)

A $- 3$
B $- 1$
C $5$
D $17$
E $29$
F $80$

Worked solution

Answer:C

Expand the square:

f (x) = (\frac{2}{x} - \frac{1}{2 x ^{2}})^{2} = \frac{4}{x ^{2}} - \frac{2}{x ^{3}} + \frac{1}{4 x ^{4}} = 4 x^{- 2} - 2 x^{- 3} + \frac{1}{4} x^{- 4} .

Differentiate once:

f^{'} (x) = - 8 x^{- 3} + 6 x^{- 4} - x^{- 5} .

Differentiate again:

f^{''} (x) = 24 x^{- 4} - 24 x^{- 5} + 5 x^{- 6} .

Substitute

x = 1

f^{''} (1) = 24 - 24 + 5 = 5.

Hence the answer is C.

TMUA 2017 Paper 2 · Q1

Given that

y = \frac{( 1 - 3 x ) ^{2}}{2 x ^{\frac{3}{2}}}

, which one of the following is a correct expression for

\frac{d y}{d x}

A $\frac{9}{4} x^{- \frac{1}{2}} + \frac{3}{2} x^{- \frac{3}{2}} - \frac{3}{4} x^{- \frac{5}{2}}$
B $\frac{9}{4} x^{- \frac{1}{2}} - \frac{3}{2} x^{- \frac{3}{2}} + \frac{3}{4} x^{- \frac{5}{2}}$
C $\frac{9}{4} x^{- \frac{1}{2}} - \frac{3}{2} x^{- \frac{3}{2}} - \frac{3}{4} x^{- \frac{5}{2}}$
D $- \frac{9}{4} x^{- \frac{1}{2}} + \frac{3}{2} x^{- \frac{3}{2}} + \frac{3}{4} x^{- \frac{5}{2}}$
E $- \frac{9}{4} x^{- \frac{1}{2}} + \frac{3}{2} x^{- \frac{3}{2}} - \frac{3}{4} x^{- \frac{5}{2}}$
F $- \frac{9}{4} x^{- \frac{1}{2}} - \frac{3}{2} x^{- \frac{3}{2}} - \frac{3}{4} x^{- \frac{5}{2}}$

Worked solution

Answer:A

Expand the numerator:

(1 - 3 x)^{2} = 1 - 6 x + 9 x^{2}

. Then

y = \frac{1 - 6 x + 9 x ^{2}}{2 x ^{\frac{3}{2}}} = \frac{1}{2} x^{- \frac{3}{2}} - 3 x^{- \frac{1}{2}} + \frac{9}{2} x^{\frac{1}{2}} .

Differentiating term by term:

\frac{d y}{d x} = \frac{1}{2} \cdot (- \frac{3}{2}) x^{- \frac{5}{2}} - 3 \cdot (- \frac{1}{2}) x^{- \frac{3}{2}} + \frac{9}{2} \cdot \frac{1}{2} x^{- \frac{1}{2}} = - \frac{3}{4} x^{- \frac{5}{2}} + \frac{3}{2} x^{- \frac{3}{2}} + \frac{9}{4} x^{- \frac{1}{2}} .

This matches option A. Hence the answer is A.