Short Questions (Question 8 – 10)

Question 8:

Find

\frac{d s}{d t}

for each of the following functions.

\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{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{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 9:

Given that y = \frac{1 - 5 x^{4}}{x - 3}, find \frac{d y}{d x} .

Solution:

\begin{array}{l} \frac{d y}{d x} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^{2}} = \frac{(x - 3) . - 20 x^{3} - (1 - 5 x^{4}) .1}{{(x - 3)}^{2}} \\ \frac{d y}{d x} = \frac{- 20 x^{4} + 60 x^{3} - 1 + 5 x^{4}}{{(x - 3)}^{2}} \\ \frac{d y}{d x} = \frac{- 15 x^{4} + 60 x^{3} - 1}{{(x - 3)}^{2}} \end{array}

Question 10:

Given that f (x) = \frac{{(x^{2} - 3)}^{5}}{1 - 3 x}, find f' (0) .

Solution:

\begin{array}{l} f (x) = \frac{{(x^{2} - 3)}^{5}}{1 - 3 x} \\ f' (x) = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^{2}} = \frac{(1 - 3 x) .5 {(x^{2} - 3)}^{4} .2 x - {(x^{2} - 3)}^{5} . - 3}{{(1 - 3 x)}^{2}} \\ f' (x) = \frac{10 x (1 - 3 x) {(x^{2} - 3)}^{4} + 3 {(x^{2} - 3)}^{5}}{{(1 - 3 x)}^{2}} \\ f' (x) = \frac{{(x^{2} - 3)}^{4} [10 x - 30 x^{2} + 3 (x^{2} - 3)]}{{(1 - 3 x)}^{2}} \\ f' (x) = \frac{{(x^{2} - 3)}^{4} [- 27 x^{2} + 10 x - 9]}{{(1 - 3 x)}^{2}} \\ ∴ f' (0) = \frac{{(0^{2} - 3)}^{4} [- 27 {(0)}^{2} + 10 (0) - 9]}{{(1 - 3 (0))}^{2}} \\ f' (0) = \frac{81 \times (- 9)}{1} = - 729 \end{array}