 # 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
$\frac{d}{dx}$ (8x3 + 4x2 – 10x)
= 24x + 8x –10

Question 2:
.

Solution:
$\begin{array}{l}y=\frac{{x}^{3}+2{x}^{2}+1}{3x}\\ y=\frac{{x}^{3}}{3x}+\frac{2{x}^{2}}{3x}+\frac{1}{3x}\\ y=\frac{{x}^{2}}{3}+\frac{2x}{3}+\frac{1}{3}{x}^{-1}\\ \frac{dy}{dx}=\frac{2x}{3}+\frac{2}{3}-\frac{1}{3}{x}^{-2}\\ \frac{dy}{dx}=\frac{2x}{3}+\frac{2}{3}-\frac{1}{3{x}^{2}}\end{array}$

Question 3:

Solution:
$\begin{array}{l}y=\frac{3}{\sqrt{5x+1}}=3{\left(5x+1\right)}^{-\frac{1}{2}}\\ \frac{dy}{dx}=-\frac{1}{2}.3{\left(5x+1\right)}^{-\frac{3}{2}}\left(5\right)\\ \frac{dy}{dx}=\frac{-15}{2{\left[{\left(5x+1\right)}^{3}\right]}^{\frac{1}{2}}}\\ \frac{dy}{dx}=\frac{-15}{2\sqrt{{\left(5x+1\right)}^{3}}}\end{array}$

Question 4:
Find $\frac{ds}{dt}$ for each of the following functions.

Solution:
(a)
$\begin{array}{l}s={\left(t-\frac{3}{t}\right)}^{2}\\ s=\left(t-\frac{3}{t}\right)\left(t-\frac{3}{t}\right)\\ s={t}^{2}-6+\frac{9}{{t}^{2}}\\ s={t}^{2}-6+9{t}^{-2}\\ \frac{ds}{dt}=2t-18{t}^{-3}=2t-\frac{18}{{t}^{3}}\end{array}$

(b)
$\begin{array}{l}s=\frac{\left(t+1\right)\left(3-5t\right)}{{t}^{2}}\\ s=\frac{3t-5{t}^{2}+3-5t}{{t}^{2}}=\frac{-5{t}^{2}-2t+3}{{t}^{2}}\\ s=-5-\frac{2}{t}+\frac{3}{{t}^{2}}=-5-2{t}^{-1}+3{t}^{-2}\\ \frac{ds}{dt}=2{t}^{-2}-6{t}^{-3}=\frac{2}{{t}^{2}}-\frac{6}{{t}^{3}}\end{array}$

Question 5:
Given that  $y=\frac{3}{5}{u}^{5}$ , where u = 4+ 1. Find $\frac{dy}{dx}$ in terms of x.

Solution: