2.10.2 Differentiation Short Questions (Question 6 - 10)

Question 6:

Given that f (x) = 3x²(4x²– 1)⁷, find f’(x).

Solution:

f (x) = 3x²(4x²– 1)⁷

f’(x) = 3x². 7(4x²– 1)⁶. 8x + (4x²– 1)⁷. 6x

f’(x) = 168x³ (4x²– 1)⁶ + 6x (4x²– 1)⁷

f’(x) = 6x (4x²– 1)⁶ [28x²+ (4x²– 1)]

f’(x) = 6x (4x²– 1)⁶ (32x²– 1)

Question 7:

Given that y = (1 + 4x)³(3x²– 1)⁴, find

\frac{d y}{d x}

$\frac{d y}{d x}$ .

Solution:

y = (1 + 4x)³(3x²– 1)⁴

\frac{d y}{d x}

$\frac{d y}{d x}$

= (1 + 4x)³. 4(3x²– 1)³.6x + (3x²– 1)⁴. 3(1 + 4x)².4

= 24x (1 + 4x)³(3x²– 1)³ + 12 (3x² – 1)⁴(1 + 4x)²

= 12 (1 + 4x)²(3x²– 1)³ [2x (1 + 4x) + (3x²– 1)]

= 12 (1 + 4x)²(3x²– 1)³ [2x + 8x² + 3x²– 1]

= 12 (1 + 4x)²(3x²– 1)³ [11x² + 2x – 1]

Question 8:

Given that f (x) = 3 x \sqrt{4 x^{2} - 1}, find f ‘ (x) .

$Given that f (x) = 3 x \sqrt{4 x^{2} - 1}, find f ‘ (x) .$ .

Solution:

\begin{matrix} f (x) = 3 x \sqrt{4 x^{2} - 1} = 3 x {(4 x^{2} - 1)}^{\frac{1}{2}} f ‘ (x) = 3 x . \frac{1}{2} {(4 x^{2} - 1)}^{- \frac{1}{2}} .8 x + {(4 x^{2} - 1)}^{\frac{1}{2}} .3 f ‘ (x) = 12 x^{2} {(4 x^{2} - 1)}^{- \frac{1}{2}} + 3 {(4 x^{2} - 1)}^{\frac{1}{2}} f ‘ (x) = 3 {(4 x^{2} - 1)}^{- \frac{1}{2}} [4 x^{2} + (4 x^{2} - 1)] f ‘ (x) = \frac{3 (8 x^{2} - 1)}{\sqrt{(4 x^{2} - 1)}} \end{matrix}

$\begin{array}{l} f (x) = 3 x \sqrt{4 x^{2} - 1} = 3 x {(4 x^{2} - 1)}^{\frac{1}{2}} \\ f ‘ (x) = 3 x . \frac{1}{2} {(4 x^{2} - 1)}^{- \frac{1}{2}} .8 x + {(4 x^{2} - 1)}^{\frac{1}{2}} .3 \\ f ‘ (x) = 12 x^{2} {(4 x^{2} - 1)}^{- \frac{1}{2}} + 3 {(4 x^{2} - 1)}^{\frac{1}{2}} \\ f ‘ (x) = 3 {(4 x^{2} - 1)}^{- \frac{1}{2}} [4 x^{2} + (4 x^{2} - 1)] \\ f ‘ (x) = \frac{3 (8 x^{2} - 1)}{\sqrt{(4 x^{2} - 1)}} \end{array}$

Question 9:

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

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

Solution:

\begin{matrix} \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{matrix}

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

Question 10:

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

Solution:

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

2.10.2 Differentiation Short Questions (Question 6 – 10)

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