A-Level Edexcel Maths Pure Transformations of Functions: 1. (a) Find the value of \( \fra

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 3 - 2009 - Paper 2

Question 3

$1.-(a)-Find-the-value-of-$-\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 3-2009-Paper 2.png$

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 3 - 2009 - Paper 2

Step 1

Find the value of $ \frac{dy}{dx} $ at the point where $ x = 2 $

96%

114 rated

Answer

To find ( \frac{dy}{dx} ) for the equation ( y = x^2 \sqrt{5x - 1} ), we will use the product rule and chain rule of differentiation:

Differentiate ( \sqrt{5x - 1} ) using the chain rule: [ \frac{d}{dx}(\sqrt{5x - 1}) = \frac{1}{2(5x - 1)^{1/2}} \cdot 5 = \frac{5}{2\sqrt{5x - 1}} ]
Using the product rule for differentiation: [ \frac{dy}{dx} = \frac{d}{dx}(x^2) \cdot \sqrt{5x - 1} + x^2 \cdot \frac{d}{dx}(\sqrt{5x - 1}) ] [ = 2x \sqrt{5x - 1} + x^2 \cdot \frac{5}{2\sqrt{5x - 1}} ]
Now substitute ( x = 2 ) to find the value: [ \frac{dy}{dx} \bigg|_{x=2} = 2(2) \sqrt{5(2) - 1} + (2^2) \cdot \frac{5}{2\sqrt{5(2) - 1}} ]
Calculate ( 5(2) - 1 = 9 ) so ( \sqrt{9} = 3 ): [ = 4 \cdot 3 + 4 \cdot \frac{5}{2 \cdot 3} = 12 + \frac{20}{6} = 12 + \frac{10}{3} = \frac{46}{3} ]

Thus, ( \frac{dy}{dx} ) at the point where ( x = 2 ) is ( \frac{46}{3} ) or approximately 15.3.

Step 2

Differentiate $ \frac{\sin 2x}{x^2} $ with respect to $ x $

99%

104 rated

Answer

We will use the quotient rule for differentiation:

Let ( u = \sin 2x ) and ( v = x^2 ). The quotient rule states: [ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} ]
First compute ( \frac{du}{dx} ): [ \frac{du}{dx} = \frac{d}{dx}(\sin 2x) = 2 \cos 2x ]
Now compute ( \frac{dv}{dx} ): [ \frac{dv}{dx} = \frac{d}{dx}(x^2) = 2x ]
Substitute ( u ), ( v ), ( \frac{du}{dx} ), and ( \frac{dv}{dx} ) into the quotient rule: [ \frac{d}{dx} \left( \frac{\sin 2x}{x^2} \right) = \frac{x^2(2 \cos 2x) - \sin 2x(2x)}{(x^2)^2} ]
Simplifying further results in: [ = \frac{2x^2 \cos 2x - 2x \sin 2x}{x^4} = \frac{2 \cos 2x - 2 \frac{\sin 2x}{x}}{x^3} ]

Thus, the derivative is: [ \frac{2 \cos 2x - 2 \frac{\sin 2x}{x}}{x^3} ]

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 3 - 2009 - Paper 2

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 3 - 2009 - Paper 2

Find the value of \( \frac{dy}{dx} \) at the point where \( x = 2 \)

Differentiate \( \frac{\sin 2x}{x^2} \) with respect to \( x \)

Join the A-Level students using SimpleStudy...