# Differentiation Short Questions (Question 6 - 10)

Question 6:
Given that f (x) = 3x2(4x 1)7, find f’(x).

Solution:
f (x) = 3x2(4x 1)7
f’(x) = 3x2. 7(4x 1)6. 8x + (4x 1)7. 6x
f’(x) = 168x3 (4x 1)6 + 6x (4x 1)7
f’(x) = 6x (4x 1)6 [28x2+ (4x 1)]
f’(x) = 6x (4x 1)6 (32x 1)

Question 7:
Given that y = (1 + 4x)3(3x 1)4, find $\frac{dy}{dx}$

Solution:
y = (1 + 4x)3(3x2 – 1)4
$\frac{dy}{dx}$
= (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]

Question 8:

Solution:
$\begin{array}{l}f\left(x\right)=3x\sqrt{4{x}^{2}-1}=3x{\left(4{x}^{2}-1\right)}^{\frac{1}{2}}\\ f\text{'}\left(x\right)=3x.\frac{1}{2}{\left(4{x}^{2}-1\right)}^{-\frac{1}{2}}.8x+{\left(4{x}^{2}-1\right)}^{\frac{1}{2}}.3\\ f\text{'}\left(x\right)=12{x}^{2}{\left(4{x}^{2}-1\right)}^{-\frac{1}{2}}+3{\left(4{x}^{2}-1\right)}^{\frac{1}{2}}\\ f\text{'}\left(x\right)=3{\left(4{x}^{2}-1\right)}^{-\frac{1}{2}}\left[4{x}^{2}+\left(4{x}^{2}-1\right)\right]\\ f\text{'}\left(x\right)=\frac{3\left(8{x}^{2}-1\right)}{\sqrt{\left(4{x}^{2}-1\right)}}\end{array}$

Question 9:

Solution:
$\begin{array}{l}\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{{v}^{2}}=\frac{\left(x-3\right).-20{x}^{3}-\left(1-5{x}^{4}\right).1}{{\left(x-3\right)}^{2}}\\ \frac{dy}{dx}=\frac{-20{x}^{4}+60{x}^{3}-1+5{x}^{4}}{{\left(x-3\right)}^{2}}\\ \frac{dy}{dx}=\frac{-15{x}^{4}+60{x}^{3}-1}{{\left(x-3\right)}^{2}}\end{array}$

Question 10:

Solution:
$\begin{array}{l}f\left(x\right)=\frac{{\left({x}^{2}-3\right)}^{5}}{1-3x}\\ f\text{'}\left(x\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{{v}^{2}}=\frac{\left(1-3x\right).5{\left({x}^{2}-3\right)}^{4}.2x-{\left({x}^{2}-3\right)}^{5}.-3}{{\left(1-3x\right)}^{2}}\\ f\text{'}\left(x\right)=\frac{10x\left(1-3x\right){\left({x}^{2}-3\right)}^{4}+3{\left({x}^{2}-3\right)}^{5}}{{\left(1-3x\right)}^{2}}\\ f\text{'}\left(x\right)=\frac{{\left({x}^{2}-3\right)}^{4}\left[10x-30{x}^{2}+3\left({x}^{2}-3\right)\right]}{{\left(1-3x\right)}^{2}}\\ f\text{'}\left(x\right)=\frac{{\left({x}^{2}-3\right)}^{4}\left[-27{x}^{2}+10x-9\right]}{{\left(1-3x\right)}^{2}}\\ \\ \therefore f\text{'}\left(0\right)=\frac{{\left({0}^{2}-3\right)}^{4}\left[-27{\left(0\right)}^{2}+10\left(0\right)-9\right]}{{\left(1-3\left(0\right)\right)}^{2}}\\ f\text{'}\left(0\right)=\frac{81×\left(-9\right)}{1}=-729\end{array}$