2.7.2 Pembezaan, SPM Praktis (Kertas 1)

Soalan 6:

Diberi f (x) = 3x²(4x²– 1)⁷, cari f’(x).

Penyelesaian:

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)

Soalan 7:

Diberi y = (1 + 4x)³(3x² – 1)⁴, cari dy/dx.

Penyelesaian:

y = (1 + 4x)³(3x²– 1)⁴

dy/dx

= (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]

Soalan 8:

Diberi f (x) = 3 x \sqrt{4 x^{2} - 1}, cari f ‘ (x) .

Penyelesaian:

\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}

Soalan 9:

Diberi y = \frac{1 - 5 x^{4}}{x - 3}, cari \frac{d y}{d x} .

Penyelesaian:

\begin{array}{l} \frac{d y}{d x} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^{2}} \\ \frac{d y}{d x} = \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}

Soalan 10:

Diberi f (x) = \frac{{(x^{2} - 3)}^{5}}{1 - 3 x}, cari f ‘ (0) .

Penyelesaian:

\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}} \\ f ‘ (x) = \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}