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Given $y = 2x(x^2 - 1)^5$, show that you need to find \(\frac{dy}{dx}\) = g(x)(x^2 - 1)^4 where g(x) is a function to be determined - Edexcel - A-Level Maths Pure - Question 8 - 2017 - Paper 4
Question 8
Given $y = 2x(x^2 - 1)^5$, show that you need to find \(\frac{dy}{dx}\) = g(x)(x^2 - 1)^4 where g(x) is a function to be determined. (a) (b) Hence find the set of... show full transcript
Worked Solution & Example Answer:Given $y = 2x(x^2 - 1)^5$, show that you need to find \(\frac{dy}{dx}\) = g(x)(x^2 - 1)^4 where g(x) is a function to be determined - Edexcel - A-Level Maths Pure - Question 8 - 2017 - Paper 4
Step 1
Show that \(\frac{dy}{dx} = g(x)(x^2 - 1)^4\)
Answer
To differentiate the given function, apply the product rule. Let (u = 2x) and (v = (x^2 - 1)^5):
- Differentiate (u): (\frac{du}{dx} = 2).
- Differentiate (v) using the chain rule: (\frac{dv}{dx} = 5(x^2 - 1)^4(2x) = 10x(x^2 - 1)^4).
- Applying the product rule: [\frac{dy}{dx} = \frac{du}{dx}v + u\frac{dv}{dx} = 2(x^2 - 1)^5 + 2x(10x(x^2 - 1)^4)]
- Factor out ((x^2 - 1)^4): [\frac{dy}{dx} = (x^2 - 1)^4(2(x^2 - 1) + 20x^2)]
- Simplifying: [\frac{dy}{dx} = g(x)(x^2 - 1)^4] where (g(x) = 2(x^2 - 1) + 20x^2).
Step 2
Hence find the set of values of x for which \(\frac{dy}{dx} > 0\)
Answer
From (g(x)): 2(x^2 - 1) + 20x^2 > 0
- Set the equation: (2x^2 - 2 + 20x^2 > 0) [22x^2 - 2 > 0]
- Solve the inequality: [22x^2 > 2] [x^2 > \frac{1}{11}]
- Thus, (x > \frac{1}{\sqrt{11}}) or (x < -\frac{1}{\sqrt{11}}) provides the required set of values.
Step 3
Given \(x = \ln(\sec(2y))\), find \(\frac{dy}{dx}\) as a function of x
Answer
Using implicit differentiation:
- Differentiate both sides: [\frac{dx}{dy} = 2\sec(2y)\tan(2y)]\ Therefore, (\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{2\sec(2y)\tan(2y)})
- Substitute (y = \frac{1}{2}\ln(e^x)) back into (\frac{dy}{dx}) for simplification: The final expression gives (\frac{dy}{dx} = \frac{1}{2\sec(2y)\tan(2y)}) as the simplest form.