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1. (a) Find the value of \( \frac{dy}{dx} \) at the point where \( x = 2 \) on the curve with equation \( y = x^2 \sqrt{5x - 1} \) - Edexcel - A-Level Maths Pure - Question 2 - 2009 - Paper 2
Question 2
1. (a) Find the value of \( \frac{dy}{dx} \) at the point where \( x = 2 \) on the curve with equation \( y = x^2 \sqrt{5x - 1} \) . (b) Differentiate \( \frac{\si... show full transcript
Worked Solution & Example Answer:1. (a) Find the value of \( \frac{dy}{dx} \) at the point where \( x = 2 \) on the curve with equation \( y = x^2 \sqrt{5x - 1} \) - Edexcel - A-Level Maths Pure - Question 2 - 2009 - Paper 2
Step 1
Find the value of \( \frac{dy}{dx} \) at the point where \( x = 2 \)
Answer
To find ( \frac{dy}{dx} ), we first differentiate the function ( y = x^2 \sqrt{5x - 1} ) using the product rule:
Let ( u = x^2 ) and ( v = \sqrt{5x - 1} ).
Then, [ \frac{dy}{dx} = u'v + uv' ]
Calculating the derivatives: [ u' = 2x ]
To find ( v' ), we use the chain rule: [ v' = \frac{1}{2}(5x - 1)^{-\frac{1}{2}} \cdot 5 = \frac{5}{2 \sqrt{5x - 1}} ]
Now substituting back: [ \frac{dy}{dx} = 2x \sqrt{5x - 1} + x^2 \cdot \frac{5}{2 \sqrt{5x - 1}} ]
Next, substituting ( x = 2 ): [ \frac{dy}{dx} \bigg|_{x=2} = 2(2) \sqrt{5(2) - 1} + (2^2) \cdot \frac{5}{2 \sqrt{5(2) - 1}} ] [ = 4 \sqrt{10 - 1} + 4 \cdot \frac{5}{2 \sqrt{10 - 1}} ] [ = 4 \sqrt{9} + 10 \frac{1}{3} ] [ = 12 + 3.3333 = \frac{46}{3} ]
Step 2
Differentiate \( \frac{\sin 2x}{x^2} \) with respect to \( x \)
Answer
Using the quotient rule to differentiate ( \frac{\sin 2x}{x^2} ):
Let ( u = \sin 2x ) and ( v = x^2 ).
The quotient rule states: [ \frac{d}{dx}\left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} ]
Calculating the derivatives: [ u' = 2 \cos 2x ] [ v' = 2x ]
Substituting back into the quotient rule: [ \frac{d}{dx}\left( \frac{\sin 2x}{x^2} \right) = \frac{(2 \cos 2x) x^2 - \sin 2x (2x)}{(x^2)^2} ] [ = \frac{2x^2 \cos 2x - 2x \sin 2x}{x^4} ] [ = \frac{2 \cos 2x - 2 \frac{\sin 2x}{x}}{x^2} ]