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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{\... 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 apply the product rule. The equation is given as:
[ y = x^2 \sqrt{5x - 1} ]
Using the product rule: [ \frac{dy}{dx} = \frac{d}{dx}(x^2) \cdot \sqrt{5x - 1} + x^2 \cdot \frac{d}{dx}(\sqrt{5x - 1}) ]
Calculating each part: [ \frac{d}{dx}(x^2) = 2x ]
For the second part, we differentiate ( \sqrt{5x - 1} ) using the chain rule: [ \frac{d}{dx}(\sqrt{5x - 1}) = \frac{1}{2\sqrt{5x - 1}} \cdot 5 = \frac{5}{2\sqrt{5x - 1}} ]
Now combine these: [ \frac{dy}{dx} = 2x \cdot \sqrt{5x - 1} + x^2 \cdot \frac{5}{2\sqrt{5x - 1}} ]
Substituting ( x = 2 ): [ \frac{dy}{dx} = 2(2) \cdot \sqrt{5(2) - 1} + (2^2) \cdot \frac{5}{2\sqrt{5(2) - 1}} ] [ = 4 \cdot \sqrt{10 - 1} + 4 \cdot \frac{5}{2\sqrt{10 - 1}} ] [ = 4 \cdot 3 + 4 \cdot \frac{5}{6} ] [ = 12 + \frac{20}{6} = 12 + \frac{10}{3} = \frac{36}{3} + \frac{10}{3} = \frac{46}{3} ] Thus, ( \frac{dy}{dx} = \frac{46}{3} ) at ( x = 2 ).
Step 2
Differentiate \( \frac{\sin 2x}{x^2} \) with respect to \( x \)
Answer
To differentiate ( \frac{\sin 2x}{x^2} ), we use the quotient rule: [ \frac{d}{dx}(\frac{u}{v}) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} ] Where ( u = \sin 2x ) and ( v = x^2 ).
Firstly, we need to find ( \frac{du}{dx} ) and ( \frac{dv}{dx} ):
[ \frac{du}{dx} = 2\cos 2x ] (using the chain rule)
[ \frac{dv}{dx} = 2x ]
Now apply the quotient rule: [ \frac{d}{dx}(\frac{\sin 2x}{x^2}) = \frac{x^2 \cdot 2\cos 2x - \sin 2x \cdot 2x}{(x^2)^2} ] [ = \frac{2x^2\cos 2x - 2x\sin 2x}{x^4} ] [ = \frac{2x (x \cos 2x - \sin 2x)}{x^4} = \frac{2(x \cos 2x - \sin 2x)}{x^3} ]