2.10.3 Differentiation Short Questions (Question 11 - 14)

Question 11:
Given that the graph of function

$f (x) = h x^{3} + \frac{k}{x^{2}}$ has a gradient function

$f ‘ (x) = 12 x^{2} - \frac{258}{x^{3}}$ such that h and k are constants. Find the values of h and k.

Solution:

$\begin{array}{l} f (x) = h x^{3} + \frac{k}{x^{2}} = h x^{3} + k x^{- 2} \\ f ‘ (x) = 3 h x^{2} - 2 k x^{- 3} \\ f ‘ (x) = 3 h x^{2} - \frac{2 k}{x^{3}} \\ But it is given that f ‘ (x) = 12 x^{2} - \frac{258}{x^{3}} \\ Hence, by comparison, \\ 3 h = 12 and 2 k = 258 \\ h = 4 k = 129 \end{array}$

Question 12:

$\begin{array}{l} Given that y = \frac{x^{2}}{x + 3}, show that \frac{d y}{d x} = \frac{x^{2} + 6 x}{{(x + 3)}^{2}} \\ Find \frac{d^{2} y}{d x^{2}} in the simplest form . \end{array}$

Solution:

$\begin{array}{l} y = \frac{x^{2}}{x + 3} \\ \frac{d y}{d x} = \frac{(x + 3) (2 x) - x^{2} .1}{{(x + 3)}^{2}} = \frac{2 x^{2} + 6 x - x^{2}}{{(x + 3)}^{2}} \\ \frac{d y}{d x} = \frac{x^{2} + 6 x}{{(x + 3)}^{2}} (shown) \\ \frac{d^{2} y}{d x^{2}} = \frac{{(x + 3)}^{2} (2 x + 6) - (x^{2} + 6 x) .2 (x + 3)}{{(x + 3)}^{4}} \\ \frac{d^{2} y}{d x^{2}} = \frac{(x + 3) [(x + 3) (2 x + 6) - 2 (x^{2} + 6 x)]}{{(x + 3)}^{4}} \\ \frac{d^{2} y}{d x^{2}} = \frac{[2 x^{2} + 6 x + 6 x + 18 - 2 x^{2} - 12 x]}{{(x + 3)}^{3}} \\ \frac{d^{2} y}{d x^{2}} = \frac{18}{{(x + 3)}^{3}} \end{array}$

Question 13:

$If y = x^{2} + 4 x, show that x^{2} \frac{d^{2} y}{d x^{2}} - 2 x \frac{d y}{d x} + 2 y = 0.$

Solution:

$\begin{array}{l} y = x^{2} + 4 x \\ \frac{d y}{d x} = 2 x + 4 \\ \frac{d^{2} y}{d x^{2}} = 2 \\ x^{2} \frac{d^{2} y}{d x^{2}} - 2 x \frac{d y}{d x} + 2 y \\ = x^{2} (2) - 2 x (2 x + 4) + 2 (x^{2} + 4 x) \\ = 2 x^{2} - 4 x^{2} - 8 x + 2 x^{2} + 8 x \\ = 0 (Shown) \end{array}$

Question 14:

Given y = x (6 – x), express

$y \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + 18$ in terms of x in the simplest form.

Hence, find the value of x which satisfies the equation

$y \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + 18 = 0$

Solution:

$\begin{array}{l} y = x (6 - x) = 6 x - x^{2} \\ \frac{d y}{d x} = 6 - 2 x \\ \frac{d^{2} y}{d x^{2}} = - 2 \\ ∴ y \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + 18 = (6 x - x^{2}) (- 2) + x (6 - 2 x) + 18 \\ = - 12 x + 2 x^{2} + 6 x - 2 x^{2} + 18 \\ = - 6 x + 18 \\ y \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + 18 = 0 \\ - 6 x + 18 = 0 \\ x = 3 \end{array}$

2.10.3 Differentiation Short Questions (Question 11 – 14)

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