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3. (a) Show that \( \frac{dy}{dx} = \frac{A}{(x + 1)^n} \) where A and n are constants to be found - Edexcel - A-Level Maths Pure - Question 5 - 2019 - Paper 1
Question 5
3. (a) Show that \( \frac{dy}{dx} = \frac{A}{(x + 1)^n} \) where A and n are constants to be found. (b) Hence deduce the range of values for x for which \( \frac{d... show full transcript
Worked Solution & Example Answer:3. (a) Show that \( \frac{dy}{dx} = \frac{A}{(x + 1)^n} \) where A and n are constants to be found - Edexcel - A-Level Maths Pure - Question 5 - 2019 - Paper 1
Step 1
Show that \( \frac{dy}{dx} = \frac{A}{(x + 1)^n} \) where A and n are constants to be found.
Answer
To find ( \frac{dy}{dx} ) for the function ( y = \frac{5x^2 + 10x}{(x + 1)^2} ), we will use the quotient rule:
[ \frac{dy}{dx} = \frac{(x + 1)^2 (10x + 10) - (5x^2 + 10x) 2 (x + 1)}{(x + 1)^4} ]
Calculating the derivative:
- Numerator: ( (x + 1)^2 (10x + 10) - (5x^2 + 10x) 2 (x + 1) )
- Expanding: ( (10x^2 + 30x + 20) - (10x^2 + 20x) = 10x + 20 )
- Thus, ( \frac{dy}{dx} = \frac{10(x + 2)}{(x + 1)^4} )
- Therefore, setting A = 10 and n = 4, we write: ( \frac{dy}{dx} = \frac{A}{(x + 1)^n} ).
Step 2
Hence deduce the range of values for x for which \( \frac{dy}{dx} < 0 \)
Answer
From the expression ( \frac{dy}{dx} = \frac{10(x + 2)}{(x + 1)^4} ), observe that the denominator is always positive (as ( (x + 1)^4 > 0 ) for all x).
For the entire expression to be negative:
[ 10(x + 2) < 0 ]
Solving this gives:
[ x + 2 < 0 ]
[ x < -2 ]
Thus, the range of values for x for which ( \frac{dy}{dx} < 0 ) is ( x < -2 ).