2.10.1 Differentiation Short Questions (Question 1 - 5)

Question 1:

Differentiate the expression 2x (4x² + 2x – 5) with respect to x.

Solution:

2x (4x² + 2x – 5) = 8x³ + 4x²– 10x

\frac{d}{d x}

$\frac{d}{d x}$ (8x³ + 4x² – 10x)

= 24x + 8x –10

Question 2:

Given that y = \frac{x^{3} + 2 x^{2} + 1}{3 x}, find \frac{d y}{d x} .

$Given that y = \frac{x^{3} + 2 x^{2} + 1}{3 x}, find \frac{d y}{d x} .$ .

Solution:

\begin{matrix} y = \frac{x^{3} + 2 x^{2} + 1}{3 x} y = \frac{x^{3}}{3 x} + \frac{2 x^{2}}{3 x} + \frac{1}{3 x} y = \frac{x^{2}}{3} + \frac{2 x}{3} + \frac{1}{3} x^{- 1} \frac{d y}{d x} = \frac{2 x}{3} + \frac{2}{3} - \frac{1}{3} x^{- 2} \frac{d y}{d x} = \frac{2 x}{3} + \frac{2}{3} - \frac{1}{3 x^{2}} \end{matrix}

$\begin{array}{l} y = \frac{x^{3} + 2 x^{2} + 1}{3 x} \\ y = \frac{x^{3}}{3 x} + \frac{2 x^{2}}{3 x} + \frac{1}{3 x} \\ y = \frac{x^{2}}{3} + \frac{2 x}{3} + \frac{1}{3} x^{- 1} \\ \frac{d y}{d x} = \frac{2 x}{3} + \frac{2}{3} - \frac{1}{3} x^{- 2} \\ \frac{d y}{d x} = \frac{2 x}{3} + \frac{2}{3} - \frac{1}{3 x^{2}} \end{array}$

Question 3:

Given that y = \frac{3}{\sqrt{5 x + 1}}, find \frac{d y}{d x}

$Given that y = \frac{3}{\sqrt{5 x + 1}}, find \frac{d y}{d x}$

Solution:

\begin{matrix} y = \frac{3}{\sqrt{5 x + 1}} = 3 {(5 x + 1)}^{- \frac{1}{2}} \frac{d y}{d x} = - \frac{1}{2} .3 {(5 x + 1)}^{- \frac{3}{2}} (5) \frac{d y}{d x} = \frac{- 15}{2 {[{(5 x + 1)}^{3}]}^{\frac{1}{2}}} \frac{d y}{d x} = \frac{- 15}{2 \sqrt{{(5 x + 1)}^{3}}} \end{matrix}

$\begin{array}{l} y = \frac{3}{\sqrt{5 x + 1}} = 3 {(5 x + 1)}^{- \frac{1}{2}} \\ \frac{d y}{d x} = - \frac{1}{2} .3 {(5 x + 1)}^{- \frac{3}{2}} (5) \\ \frac{d y}{d x} = \frac{- 15}{2 {[{(5 x + 1)}^{3}]}^{\frac{1}{2}}} \\ \frac{d y}{d x} = \frac{- 15}{2 \sqrt{{(5 x + 1)}^{3}}} \end{array}$

Question 4:

Find

\frac{d s}{d t}

$\frac{d s}{d t}$ for each of the following functions.

\begin{matrix} (a) s = {(t - \frac{3}{t})}^{2} (b) s = \frac{(t + 1) (3 - 5 t)}{t^{2}} \end{matrix}

$\begin{array}{l} (a) s = {(t - \frac{3}{t})}^{2} \\ (b) s = \frac{(t + 1) (3 - 5 t)}{t^{2}} \end{array}$

Solution:
(a)

\begin{matrix} s = {(t - \frac{3}{t})}^{2} s = (t - \frac{3}{t}) (t - \frac{3}{t}) s = t^{2} - 6 + \frac{9}{t^{2}} s = t^{2} - 6 + 9 t^{- 2} \frac{d s}{d t} = 2 t - 18 t^{- 3} = 2 t - \frac{18}{t^{3}} \end{matrix}

$\begin{array}{l} s = {(t - \frac{3}{t})}^{2} \\ s = (t - \frac{3}{t}) (t - \frac{3}{t}) \\ s = t^{2} - 6 + \frac{9}{t^{2}} \\ s = t^{2} - 6 + 9 t^{- 2} \\ \frac{d s}{d t} = 2 t - 18 t^{- 3} = 2 t - \frac{18}{t^{3}} \end{array}$

(b)

\begin{matrix} s = \frac{(t + 1) (3 - 5 t)}{t^{2}} s = \frac{3 t - 5 t^{2} + 3 - 5 t}{t^{2}} = \frac{- 5 t^{2} - 2 t + 3}{t^{2}} s = - 5 - \frac{2}{t} + \frac{3}{t^{2}} = - 5 - 2 t^{- 1} + 3 t^{- 2} \frac{d s}{d t} = 2 t^{- 2} - 6 t^{- 3} = \frac{2}{t^{2}} - \frac{6}{t^{3}} \end{matrix}

$\begin{array}{l} s = \frac{(t + 1) (3 - 5 t)}{t^{2}} \\ s = \frac{3 t - 5 t^{2} + 3 - 5 t}{t^{2}} = \frac{- 5 t^{2} - 2 t + 3}{t^{2}} \\ s = - 5 - \frac{2}{t} + \frac{3}{t^{2}} = - 5 - 2 t^{- 1} + 3 t^{- 2} \\ \frac{d s}{d t} = 2 t^{- 2} - 6 t^{- 3} = \frac{2}{t^{2}} - \frac{6}{t^{3}} \end{array}$

Question 5:

Given that

y = \frac{3}{5} u^{5}

$y = \frac{3}{5} u^{5}$ , where u = 4x + 1. Find

\frac{d y}{d x}

$\frac{d y}{d x}$ in terms of x.

Solution:

\begin{matrix} y = \frac{3}{5} u^{5}, u = 4 x + 1 y = \frac{3}{5} {(4 x + 1)}^{5} \frac{d y}{d x} = 5. \frac{3}{5} {(4 x + 1)}^{4} .4 \frac{d y}{d x} = 12 {(4 x + 1)}^{4} \end{matrix}

$\begin{array}{l} y = \frac{3}{5} u^{5}, u = 4 x + 1 \\ y = \frac{3}{5} {(4 x + 1)}^{5} \\ \frac{d y}{d x} = 5. \frac{3}{5} {(4 x + 1)}^{4} .4 \\ \frac{d y}{d x} = 12 {(4 x + 1)}^{4} \end{array}$

2.10.1 Differentiation Short Questions (Question 1 – 5)

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