The curve C has equation $$y = \frac{(x^2 + 4)(x - 3)}{2x}, \quad x \neq 0$$ (a) Find $ \frac{dy}{dx} $ in its simplest form. (b) Find an equation of the tangent to C at the point where $ x = -1 $ Give your answer in the form $ ax + by + c = 0 $, where a, b and c are integers.

A-Level Edexcel Maths Pure Solving Equations: The curve C has equation $$y =

The curve C has equation $$y = \frac{(x^2 + 4)(x - 3)}{2x}, \quad x \neq 0$$ (a) Find $ \frac{dy}{dx} $ in its simplest form - Edexcel - A-Level Maths Pure - Question 7 - 2015 - Paper 1

Question 7

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The curve C has equation $$y = \frac{(x^2 + 4)(x - 3)}{2x}, \quad x \neq 0$$ (a) Find $ \frac{dy}{dx} $ in its simplest form. (b) Find an equation of the tangen... show full transcript

Worked Solution & Example Answer:The curve C has equation $$y = \frac{(x^2 + 4)(x - 3)}{2x}, \quad x \neq 0$$ (a) Find $ \frac{dy}{dx} $ in its simplest form - Edexcel - A-Level Maths Pure - Question 7 - 2015 - Paper 1

Step 1

Find $ \frac{dy}{dx} $ in its simplest form.

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Answer

To find ( \frac{dy}{dx} ), we will first apply the product rule and then simplify the expression.

Rewrite the equation as: $y = \frac{(x^2 + 4)(x - 3)}{2x}$
Identify (u = (x^2 + 4)(x - 3)) and (v = 2x).
Use the quotient rule: $\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$
Calculate (\frac{du}{dx}) and (\frac{dv}{dx}):
- (\frac{du}{dx} = (x^2 + 4) + (x - 3)(2x) = 3x^2 - 3 + 4 = 3x^2 + 1)
- (\frac{dv}{dx} = 2)
Substitute these into the quotient rule: $\frac{dy}{dx} = \frac{2x(3x^2 + 1) - (x^2 + 4)(x - 3)(2)}{(2x)^2}$
Simplify the expression to obtain (\frac{dy}{dx} ) in its simplest form.

Step 2

Find an equation of the tangent to C at the point where $ x = -1 $

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Answer

To find the equation of the tangent at ( x = -1 ):

First, we find the value of ( y ) when ( x = -1 ): $y = \frac{((-1)^2 + 4)((-1) - 3)}{2(-1)} = \frac{(1 + 4)(-4)}{-2} = \frac{5(-4)}{-2} = 10$ So, the point is ((-1, 10)).
Next, we find ( \frac{dy}{dx} ) at ( x = -1 ) to get the slope of the tangent: Substitute ( x = -1 ) into the derived ( \frac{dy}{dx} ).
Assuming we calculated ( \frac{dy}{dx} |_{x=-1} = m ) (you will need the simplified value here).
We now use the point-slope form to determine the equation of the tangent line: $y - 10 = m(x + 1)$
Rearranging this to the form ( ax + by + c = 0 ): $\Rightarrow mx - y + 10 + m = 0$
Here, a, b, and c can now be identified as suitable integers.

The curve C has equation $$y = \frac{(x^2 + 4)(x - 3)}{2x}, \quad x \neq 0$$ (a) Find \( \frac{dy}{dx} \) in its simplest form - Edexcel - A-Level Maths Pure - Question 7 - 2015 - Paper 1

Worked Solution & Example Answer:The curve C has equation $$y = \frac{(x^2 + 4)(x - 3)}{2x}, \quad x \neq 0$$ (a) Find \( \frac{dy}{dx} \) in its simplest form - Edexcel - A-Level Maths Pure - Question 7 - 2015 - Paper 1

Find \( \frac{dy}{dx} \) in its simplest form.

Find an equation of the tangent to C at the point where \( x = -1 \)

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