A-Level Edexcel Maths Pure Graphs of Functions: The curve C has equation $$y =

The curve C has equation $$y = \frac{(x+3)(x-8)}{x}, \quad x > 0$$ (a) Find $ \frac{dy}{dx} $ in its simplest form - Edexcel - A-Level Maths Pure - Question 8 - 2010 - Paper 2

Question 8

$The-curve-C-has-equation--$$y-=-\frac{(x+3)(x-8)}{x},-\quad-x->-0$$--(a)-Find-$-\frac{dy}{dx}-$-in-its-simplest-form-Edexcel-A-Level Maths Pure-Question 8-2010-Paper 2.png$

The curve C has equation $$y = \frac{(x+3)(x-8)}{x}, \quad x > 0$$ (a) Find $ \frac{dy}{dx} $ in its simplest form. (b) Find an equation of the tangent to C at ... show full transcript

Worked Solution & Example Answer:The curve C has equation $$y = \frac{(x+3)(x-8)}{x}, \quad x > 0$$ (a) Find $ \frac{dy}{dx} $ in its simplest form - Edexcel - A-Level Maths Pure - Question 8 - 2010 - Paper 2

Step 1

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

96%

114 rated

Answer

To differentiate the given equation, we can use the quotient rule.

Given: $y = \frac{(x+3)(x-8)}{x}$

We first rewrite this as: $y = (x+3)(x-8) \cdot x^{-1}$

Applying the product rule: $\frac{dy}{dx} = \frac{d}{dx}((x+3)(x-8)) \cdot x^{-1} + (x+3)(x-8) \cdot \frac{d}{dx}(x^{-1})$

Calculating the first derivative:

Differentiate ((x+3)(x-8)):
- (= (x+3)\cdot 1 + (x-8)\cdot 1 = 2x - 5)
Differentiate (x^{-1}): (= -x^{-2})

Thus, substituting back: $\frac{dy}{dx} = (2x - 5) \cdot x^{-1} - (x+3)(x-8)x^{-2}$

This simplifies into: $\frac{dy}{dx} = \frac{2x - 5}{x} - \frac{(x+3)(x-8)}{x^2}$

Combining the fractions: $\frac{dy}{dx} = \frac{(2x-5)x - (x+3)(x-8)}{x^2}$

This simplifies to: $\frac{dy}{dx} = \frac{24 - 5x}{x^2}$

So, the simplest form is: $\frac{dy}{dx} = \frac{24 - 5x}{x^2}$

Step 2

Find an equation of the tangent to C at the point where $ x = 2 $

99%

104 rated

Answer

First, we need to find the gradient at the point where ( x = 2 ) using our earlier derivative:

( \frac{dy}{dx} = \frac{24 - 5x}{x^2} )

For ( x = 2 ): ( \frac{dy}{dx} = \frac{24 - 5(2)}{2^2} = \frac{14}{4} = \frac{7}{2} ).

Now we need the y-coordinate when ( x = 2 ):

( y = \frac{(2+3)(2-8)}{2} = \frac{5(-6)}{2} = -15 ).

The point of tangency is (2, -15). Using point-slope form for the line, the equation of the tangent is:

( y - y_1 = m(x - x_1) ) where ( m = \frac{7}{2} ) and ( (x_1, y_1) = (2, -15) ):

( y + 15 = \frac{7}{2}(x - 2) ).

This simplifies to: ( y = \frac{7}{2}x - 7 - 15 ), ( y = \frac{7}{2}x - 22 ).

Thus, the equation of the tangent is: ( y = \frac{7}{2}x - 22 ).

The curve C has equation $$y = \frac{(x+3)(x-8)}{x}, \quad x > 0$$ (a) Find \( \frac{dy}{dx} \) in its simplest form - Edexcel - A-Level Maths Pure - Question 8 - 2010 - Paper 2

Worked Solution & Example Answer:The curve C has equation $$y = \frac{(x+3)(x-8)}{x}, \quad x > 0$$ (a) Find \( \frac{dy}{dx} \) in its simplest form - Edexcel - A-Level Maths Pure - Question 8 - 2010 - Paper 2

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

Find an equation of the tangent to C at the point where \( x = 2 \)

Join the A-Level students using SimpleStudy...