Question

due: friday november 19, 2025 total marks: /35
**3 communication marks are given for:
□ use clear, consistent, and correct mathematical notation (e.g., ( f(x),\frac{dy}{dx} ), etc.).
□ presents solutions in a logical order with correct symbols and terminology related to derivatives (e.g., derivative rules,
differentiability, tangent line concepts)
□ for applied problems, communicates the meaning of the derivative in context (e.g., the derivative represents the
velocity of the particle, or the slope of the tangent line indicates the rate at which cost is increasing.).
/3
knowledge (12 marks)

1. differentiate. simplify your answer.

a) ( y=sqrt{2x}-x^{pi}+\frac{1}{x - 1} ) 2 marks b) ( y=(2x^{3}-pi)^{2}(x^{3}-1)^{3} ) 3 marks
c) ( y=left(\frac{3x + 1}{2x-1}
ight)^{4} ) 3 marks d) ( y=\frac{5x + 1}{sqrt{3-4x}} ) 4 marks

Explanation:

Step1: Differentiate \(y = \sqrt{2x}-x^{\pi}+\frac{1}{x - 1}\)

Rewrite \(\sqrt{2x}=(2x)^{\frac{1}{2}}\), \(\frac{1}{x - 1}=(x - 1)^{-1}\).
Using the power rule \((x^n)^\prime=nx^{n - 1}\), \((ax^n)^\prime=anx^{n - 1}\) (where \(a\) is a constant).
\(y^\prime=\frac{1}{2}(2x)^{-\frac{1}{2}}\times2-\pi x^{\pi - 1}-(x - 1)^{-2}\)
Simplify: \(y^\prime=\frac{1}{\sqrt{2x}}-\pi x^{\pi - 1}-\frac{1}{(x - 1)^2}\)

Step2: Differentiate \(y=(2x^{3}-\pi)^{2}(x^{3}-1)^{3}\)

Use the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u=(2x^{3}-\pi)^{2}\) and \(v=(x^{3}-1)^{3}\).
First, find \(u^\prime\) using the chain rule \((f(g(x)))^\prime=f^\prime(g(x))\cdot g^\prime(x)\). Let \(f(u)=u^{2}\), \(g(x)=2x^{3}-\pi\), then \(u^\prime = 2(2x^{3}-\pi)\times6x^{2}=12x^{2}(2x^{3}-\pi)\)
Let \(f(u)=u^{3}\), \(g(x)=x^{3}-1\), then \(v^\prime=3(x^{3}-1)^{2}\times3x^{2}=9x^{2}(x^{3}-1)^{2}\)
\(y^\prime=12x^{2}(2x^{3}-\pi)(x^{3}-1)^{3}+9x^{2}(2x^{3}-\pi)^{2}(x^{3}-1)^{2}\)
Factor out \(3x^{2}(2x^{3}-\pi)(x^{3}-1)^{2}\): \(y^\prime = 3x^{2}(2x^{3}-\pi)(x^{3}-1)^{2}[4(x^{3}-1)+3(2x^{3}-\pi)]\)
Simplify the bracket: \(y^\prime=3x^{2}(2x^{3}-\pi)(x^{3}-1)^{2}(4x^{3}-4 + 6x^{3}-3\pi)=3x^{2}(2x^{3}-\pi)(x^{3}-1)^{2}(10x^{3}-3\pi - 4)\)

Step3: Differentiate \(y = (\frac{3x + 1}{2x-1})^{4}\)

Use the chain rule and the quotient rule. Let \(u=\frac{3x + 1}{2x-1}\), \(y = u^{4}\).
Quotient rule \((\frac{f(x)}{g(x)})^\prime=\frac{f^\prime(x)g(x)-f(x)g^\prime(x)}{g(x)^{2}}\), where \(f(x)=3x + 1\), \(f^\prime(x)=3\), \(g(x)=2x-1\), \(g^\prime(x)=2\)
\(u^\prime=\frac{3(2x - 1)-2(3x + 1)}{(2x-1)^{2}}=\frac{6x-3-6x - 2}{(2x-1)^{2}}=-\frac{5}{(2x-1)^{2}}\)
By the chain rule \(y^\prime = 4u^{3}\cdot u^\prime\)
Substitute \(u\) and \(u^\prime\): \(y^\prime=4(\frac{3x + 1}{2x-1})^{3}\times(-\frac{5}{(2x-1)^{2}})=-\frac{20(3x + 1)^{3}}{(2x-1)^{5}}\)

Step4: Differentiate \(y=\frac{5x + 1}{\sqrt{3-4x}}\)

Rewrite \(\sqrt{3-4x}=(3-4x)^{\frac{1}{2}}\). Use the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v-uv^\prime}{v^{2}}\), where \(u = 5x+1\), \(u^\prime=5\), \(v=(3-4x)^{\frac{1}{2}}\)
Find \(v^\prime\) using the chain rule: \(v^\prime=\frac{1}{2}(3-4x)^{-\frac{1}{2}}\times(-4)=-\frac{2}{\sqrt{3-4x}}\)
\(y^\prime=\frac{5\sqrt{3-4x}-(5x + 1)(-\frac{2}{\sqrt{3-4x}})}{3-4x}\)
Multiply numerator and denominator by \(\sqrt{3-4x}\):
\(y^\prime=\frac{5(3-4x)+2(5x + 1)}{(3-4x)^{\frac{3}{2}}}=\frac{15-20x+10x + 2}{(3-4x)^{\frac{3}{2}}}=\frac{17-10x}{(3-4x)^{\frac{3}{2}}}\)

Answer:

a) \(y^\prime=\frac{1}{\sqrt{2x}}-\pi x^{\pi - 1}-\frac{1}{(x - 1)^2}\)
b) \(y^\prime=3x^{2}(2x^{3}-\pi)(x^{3}-1)^{2}(10x^{3}-3\pi - 4)\)
c) \(y^\prime=-\frac{20(3x + 1)^{3}}{(2x-1)^{5}}\)
d) \(y^\prime=\frac{17-10x}{(3-4x)^{\frac{3}{2}}}\)