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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... 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} ), we need to differentiate the given function ( y = \frac{5x^{2}+10x}{(x+1)^{2}} ) using the quotient rule:
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Quotient Rule: If ( y = \frac{u}{v} ), then ( \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^{2}} )
- Here, ( u = 5x^{2} + 10x ) and ( v = (x + 1)^{2} ).
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Differentiate u and v:
- ( \frac{du}{dx} = 10x + 10 )
- ( \frac{dv}{dx} = 2(x + 1) )
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Apply the quotient rule: [ \frac{dy}{dx} = \frac{(x + 1)^{2}(10x + 10) - (5x^{2} + 10x)(2(x + 1))}{((x + 1)^{2})^{2}}] Simplifying this expression will yield: [ \frac{dy}{dx} = \frac{A}{(x + 1)^{n}}]
where constants ( A ) and ( n ) can be determined by further simplification.
Step 2
Hence deduce the range of values for x for which \( \frac{dy}{dx} < 0 \).
Answer
To determine the range of values for ( x ) such that ( \frac{dy}{dx} < 0 ):
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Sign Analysis: The derivative ( \frac{dy}{dx} ) represents the slope of the function. We need to analyze when this quantity is negative.
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Finding Conditions: Since ( \frac{dy}{dx} = \frac{A}{(x + 1)^{n}} ), set ( A < 0 ).
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Identify Effective Range: The expression will be negative when:
- The numerator ( A ) is negative and ( (x + 1)^{n} ) is positive, which occurs whenever ( x + 1 > 0 ) or ( x > -1 ).
Therefore, for ( \frac{dy}{dx} < 0 ), we deduce that: [ x < -1 ] is not applicable since it contradicts the positivity condition. Thus, the valid condition is: [ x > -1 ] which suffices the inequality.