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Given $y = 2x(x^2 - 1)^5$, show that dy/dx = g(x)(x^2 - 1)^4 where g(x) is a function to be determined - Edexcel - A-Level Maths Pure - Question 7 - 2017 - Paper 4
Question 7
Given $y = 2x(x^2 - 1)^5$, show that dy/dx = g(x)(x^2 - 1)^4 where g(x) is a function to be determined. (b) Hence find the set of values of $x$ for which \(\frac{d... show full transcript
Worked Solution & Example Answer:Given $y = 2x(x^2 - 1)^5$, show that dy/dx = g(x)(x^2 - 1)^4 where g(x) is a function to be determined - Edexcel - A-Level Maths Pure - Question 7 - 2017 - Paper 4
Step 1
Show that \(\frac{dy}{dx} = g(x)(x^2 - 1)^4\)
Answer
To find (\frac{dy}{dx}), we need to apply the product rule. Let:
- (u = 2x)
- (v = (x^2 - 1)^5)
Then using the product rule, (\frac{dy}{dx} = u'v + uv').
Firstly, calculate (u' = 2).
Next, we differentiate (v = (x^2 - 1)^5) using the chain rule: [v' = 5(x^2 - 1)^4 \cdot (2x) = 10x(x^2 - 1)^4.]
Now substitute back into the product rule: [\frac{dy}{dx} = 2(x^2 - 1)^5 + 2x(10x(x^2 - 1)^4)] [\frac{dy}{dx} = 2(x^2 - 1)^4 \left( (x^2 - 1) + 10x^2 \right)] [= 2(x^2 - 1)^4 (11x^2 - 1)]
Thus, we can express the result as: [\frac{dy}{dx} = g(x)(x^2 - 1)^4] Where g(x) is given by: [g(x) = 2(11x^2 - 1)]
Step 2
Hence find the set of values of x for which \(\frac{dy}{dx} > 0\)
Answer
From the expression obtained, (\frac{dy}{dx} = 2(x^2 - 1)^4(11x^2 - 1)).
Since (2(x^2 - 1)^4) is always positive, the sign of (\frac{dy}{dx}) depends on (11x^2 - 1). We set: [11x^2 - 1 > 0 \Rightarrow 11x^2 > 1 \Rightarrow x^2 > \frac{1}{11} \Rightarrow |x| > \frac{1}{\sqrt{11}}.]
Therefore, the set of values of (x) for which (\frac{dy}{dx} > 0) is: [x < -\frac{1}{\sqrt{11}} \text{ or } x > \frac{1}{\sqrt{11}}.]
Step 3
Given \(x = \ln(\sec(2y))\), find \(\frac{dy}{dx}\)
Answer
We start with the given expression (x = \ln(\sec(2y))). Differentiating both sides with respect to (x), we get: [\frac{dx}{dx} = \frac{d}{dx}(\ln(\sec(2y)))] [1 = \frac{1}{\sec(2y)} \cdot \sec(2y)\tan(2y) \cdot \frac{d(2y)}{dx}] [1 = \sec(2y)\tan(2y) \cdot 2 \frac{dy}{dx}]
Rearranging gives: [\frac{dy}{dx} = \frac{1}{2 \sec(2y)\tan(2y)}.]
Since (\sec(2y) = e^{x}), substituting we get: [\frac{dy}{dx} = \frac{1}{2e^{x} \tan(2y)}.]