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1. (a) Find the value of \( \frac{dy}{dx} \) at the point where \( x = 2 \) on the curve with equation \[ y = x^2 \sqrt{5x - 1} \] (b) Differentiate \( \frac{\sin 2x}{x^2} \) with respect to \( x \). - Edexcel - A-Level Maths Pure - Question 1 - 2009 - Paper 2
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![1.-(a)-Find-the-value-of-\(-\frac{dy}{dx}-\)-at-the-point-where-\(-x-=-2-\)-on-the-curve-with-equation-\[-y-=-x^2-\sqrt{5x---1}-\]---(b)-Differentiate-\(-\frac{\sin-2x}{x^2}-\)-with-respect-to-\(-x-\).-Edexcel-A-Level Maths Pure-Question 1-2009-Paper 2.png](https://cdn.simplestudy.io/assets/backend/uploads/question/MathsPure_2009_2_1_QP_1_2009.jpg)
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{\sin ... 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 \[ y = x^2 \sqrt{5x - 1} \] (b) Differentiate \( \frac{\sin 2x}{x^2} \) with respect to \( x \). - Edexcel - A-Level Maths Pure - Question 1 - 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 will use the product rule.
Let ( u = x^2 ) and ( v = \sqrt{5x - 1} ).
- Differentiate:
- ( \frac{du}{dx} = 2x )
- ( \frac{dv}{dx} = \frac{1}{2 \sqrt{5x - 1}} \cdot 5 = \frac{5}{2 \sqrt{5x - 1}} )
Using the product rule:
[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} ]
Substituting for u and v, we have:
[ \frac{dy}{dx} = x^2 \left( \frac{5}{2 \sqrt{5x - 1}} \right) + \sqrt{5x - 1} \cdot 2x ]
- Evaluating at ( x = 2 ):
- Substitute ( x = 2 ) into the equation:
- ( y = 2^2 \sqrt{5(2) - 1} = 4 \sqrt{9} = 12 )
- Thus, ( \frac{dy}{dx} ) at ( x = 2 ):
[ \frac{dy}{dx} = 2^2 \left( \frac{5}{2 \cdot 3} \right) + 3 \cdot 4 = \frac{20}{6} + 12 = \frac{10}{3} + 12 = \frac{46}{3} ]
The answer is: ( \frac{46}{3} ).
Step 2
Differentiate \( \frac{\sin 2x}{x^2} \) with respect to \( x \)
Answer
To differentiate ( \frac{\sin 2x}{x^2} ), we will again use the quotient rule.
Let ( u = \sin 2x ) and ( v = x^2 ).
- Differentiate:
- ( \frac{du}{dx} = 2 \cos 2x )
- ( \frac{dv}{dx} = 2x )
Using the quotient rule:
[ \frac{d}{dx}\left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} ]
Substituting for u and v, we get:
[ \frac{d}{dx}\left( \frac{\sin 2x}{x^2} \right) = \frac{x^2 \cdot 2 \cos 2x - \sin 2x \cdot 2x}{x^4} ]
- Simplify to get:
[ \frac{2x \cos 2x - 2 \sin 2x}{x^3} ]
Thus, the final differentiated result is:
[ \frac{2 \cos 2x - 2 \sin 2x}{x^3} ].