Short Questions (Question 1 – 4)

Posted on April 22, 2020 by user

Question 1:

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

Solution:

2x (4x² + 2x – 5) = 8x³ + 4x²– 10x

$\frac{d}{d x}$ (8x³ + 4x² – 10x)

= 24x + 8x –10

Question 2:

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

Solution:

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

Question 3:

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

Solution:

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

Question 4:

Given that

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

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

Solution:

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