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y = \frac{4x}{x^2 + 5} (a) Find \frac{dy}{dx}, writing your answer as a single fraction in its simplest form - Edexcel - A-Level Maths Pure - Question 3 - 2016 - Paper 3
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y = \frac{4x}{x^2 + 5} (a) Find \frac{dy}{dx}, writing your answer as a single fraction in its simplest form. (b) Hence find the set of values of x for which \frac... show full transcript
Worked Solution & Example Answer:y = \frac{4x}{x^2 + 5} (a) Find \frac{dy}{dx}, writing your answer as a single fraction in its simplest form - Edexcel - A-Level Maths Pure - Question 3 - 2016 - Paper 3
Step 1
Find \frac{dy}{dx}, writing your answer as a single fraction in its simplest form.
Answer
To find ( \frac{dy}{dx} ), we apply the quotient rule:
Given ( y = \frac{u}{v} ) where ( u = 4x ) and ( v = x^2 + 5 ), the quotient rule states that:
[ \frac{dy}{dx} = \frac{u'v - uv'}{v^2} ]
Calculating ( u' ) and ( v' ):
- ( u' = 4 )
- ( v' = 2x )
Substituting these derivatives into the quotient rule:
[ \frac{dy}{dx} = \frac{(4)(x^2 + 5) - (4x)(2x)}{(x^2 + 5)^2} ]
This simplifies to:
[ \frac{dy}{dx} = \frac{4x^2 + 20 - 8x^2}{(x^2 + 5)^2} = \frac{-4x^2 + 20}{(x^2 + 5)^2} ]
Therefore, the simplest form of ( \frac{dy}{dx} ) is:
[ \frac{dy}{dx} = \frac{20 - 4x^2}{(x^2 + 5)^2} ]
Step 2
Hence find the set of values of x for which \frac{dy}{dx} < 0.
Answer
To determine when ( \frac{dy}{dx} < 0 ), we must analyze:
[ \frac{20 - 4x^2}{(x^2 + 5)^2} < 0 ]
The denominator, ( (x^2 + 5)^2 ), is always positive, so the sign of ( \frac{dy}{dx} ) is determined by the numerator:
[ 20 - 4x^2 < 0 ]
This can be rearranged to:
[ 4x^2 > 20 ] [ x^2 > 5 ]
Taking the square root of both sides results in:
[ |x| > \sqrt{5} ]
Thus, the solution set is:
[ x < -\sqrt{5} \quad \text{or} \quad x > \sqrt{5} ]