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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 - Edexcel - A-Level Maths Pure - Question 8 - 2015 - Paper 1
Question 8
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 tang... 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 8 - 2015 - Paper 1
Step 1
Find \( \frac{dy}{dx} \) in its simplest form.
Answer
To find ( \frac{dy}{dx} ), we will use the quotient rule, which states that if ( y = \frac{u}{v} ), then ( \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^{2}} ).
In our case:
- ( u = (x^{2} + 4)(x - 3) )
- ( v = 2x )
First, we find ( \frac{du}{dx} ): Using the product rule: [ \frac{du}{dx} = (x^{2} + 4) \frac{d}{dx}(x - 3) + (x - 3) \frac{d}{dx}(x^{2} + 4) = (x^{2} + 4)(1) + (x - 3)(2x) = ] [ (x^{2} + 4) + (2x^{2} - 6x) = 3x^{2} - 6x + 4 ]
Next, we calculate ( \frac{dv}{dx} ): [ \frac{dv}{dx} = 2 ]
Now substituting these values back to the quotient rule: [ \frac{dy}{dx} = \frac{2x(3x^{2} - 6x + 4) - (x^{2} + 4)(2)}{(2x)^{2}} = \frac{6x^{3} - 12x^{2} + 8x - 2x^{2} - 8}{4x^{2}} ] [ = \frac{6x^{3} - 14x^{2} + 8x - 8}{4x^{2}} = \frac{2(3x^{3} - 7x^{2} + 4x - 4)}{4x^{2}} = \frac{3x^{3} - 7x^{2} + 4x - 4}{2x^{2}}\text{ in its simplest form.} ]
Step 2
Find an equation of the tangent to C at the point where \( x = -1 \)
Answer
First, we need to find the point on the curve when ( x = -1 ): [ y = \frac{((-1)^{2} + 4)((-1) - 3)}{2(-1)} = \frac{(1 + 4)(-4)}{-2} = \frac{-20}{-2} = 10 ] Thus, the point is ( (-1, 10) ).
Next, we need to find the slope of the tangent at that point. We substitute ( x = -1 ) in our derivative: [ \frac{dy}{dx} |_{x=-1} = \frac{3(-1)^{3} - 7(-1)^{2} + 4(-1) - 4}{2(-1)^{2}} = \frac{-3 - 7 - 4 - 4}{2} = \frac{-18}{2} = -9 ] Thus, the slope ( m = -9 ).
Using point-slope form of a line: [ y - y_1 = m(x - x_1) ] Substituting the point and the slope: [ y - 10 = -9(x + 1) ] Expanding: [ y - 10 = -9x - 9 ] [ y = -9x + 1 ] Rearranging into the form ( ax + by + c = 0 ): [ 9x + y - 1 = 0 ] Thus, a = 9, b = 1, and c = -1.