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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
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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. (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} ), we will differentiate the given function:
[ y = \frac{5x^2 + 10x}{(x + 1)^2} ]
We can apply the quotient rule for differentiation, which states that 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 ).
First, we find ( \frac{du}{dx} ):
[ \frac{du}{dx} = 10x + 10 ]
Next, we find ( \frac{dv}{dx} ):
[ \frac{dv}{dx} = 2(x + 1) ]
Now, substituting these into the quotient rule gives us:
[ \frac{dy}{dx} = \frac{(x + 1)^2(10x + 10) - (5x^2 + 10x)(2(x + 1))}{(x + 1)^4} ]
Simplifying the numerator:
[ = (x + 1)^2(10x + 10) - (10x^2 + 20x) = 10x^3 + 30x^2 + 20x - 10x^2 - 20x = 10x^3 + 10x^2 ]
Thus, we have:
[ \frac{dy}{dx} = \frac{10x^3 + 10x^2}{(x + 1)^4} = \frac{10x^2(x + 1)}{(x + 1)^4} = \frac{10x^2}{(x + 1)^3} ]
Therefore, we can express this as:
[ \frac{dy}{dx} = \frac{A}{(x + 1)^3} ]
Where A = 10x^2, and n = 3.
Step 2
Hence deduce the range of values for x for which \( \frac{dy}{dx} < 0 \)
Answer
From our previous result, we have:
[ \frac{dy}{dx} = \frac{10x^2}{(x + 1)^3} ]
For ( \frac{dy}{dx} < 0 ), the numerator must be negative since the denominator is positive when ( x > -1 ). Therefore, we need to ensure:
[ 10x^2 < 0 ]
However, since ( 10x^2 ) is always non-negative (as it is a square multiplied by a positive constant), ( \frac{dy}{dx} < 0 ) is not achievable unless the entire expression is set up differently. The only time it can be zero is when x = 0 or x < -1.
Thus, the conclusion is that:
[ \frac{dy}{dx} < 0 \text{ for } x < -1 ]