A-Level Edexcel Maths Pure Trigonometric Functions: 3. (a) Show that \( \frac{dy}{d

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 4 - 2019 - Paper 2

Question 4

$3.-(a)-Show-that-$-\frac{dy}{dx}-=-\frac{A}{(x-+-1)^{n}}-$-where-A-and-n-are-constants-to-be-found-Edexcel-A-Level Maths Pure-Question 4-2019-Paper 2.png$

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 4 - 2019 - Paper 2

Step 1

Show that $ \frac{dy}{dx} = \frac{A}{(x + 1)^{n}} $ where A and n are constants to be found.

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Answer

To find ( \frac{dy}{dx} ), we start with the function:

$y = \frac{5x^{2} + 10x}{(x + 1)^{2}}$

Using the quotient rule for differentiation:

$\frac{dy}{dx} = \frac{(x+1)^{2} \cdot (10x + 10) - (5x^{2} + 10x) \cdot 2(x + 1)}{((x + 1)^{2})^{2}}$

We will simplify the numerator:

Expanding the first term: ((x + 1)^{2} \cdot (10x + 10) = 10x^{3} + 30x^{2} + 30x)
Expanding the second term: ((5x^{2} + 10x) \cdot 2(x + 1) = 10x^{3} + 20x^{2})

Now substituting back gives us:

$\frac{dy}{dx} = \frac{(10x^{3} + 30x^{2} + 30x) - (10x^{3} + 20x^{2})}{(x + 1)^{4}}$

This simplifies to:

$\frac{dy}{dx} = \frac{10x^{2} + 30x}{(x + 1)^{4}}$

Factorizing the numerator results in:

$\frac{dy}{dx} = \frac{10x(x + 3)}{(x + 1)^{4}}$

Thus we can see that ( A = 10x ) and ( n = 4 ).

Step 2

Hence deduce the range of values for x for which $ \frac{dy}{dx} < 0 $

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Answer

From the previous part, we have:

$\frac{dy}{dx} = \frac{10x(x + 3)}{(x + 1)^{4}}$

To find when ( \frac{dy}{dx} < 0 ), we check the sign of the numerator since the denominator is always positive for ( x \neq -1 ).

The critical points from the numerator are where:

( 10x = 0 ) ( \Rightarrow x = 0 )
( x + 3 = 0 ) ( \Rightarrow x = -3 )

We have critical points at ( x = -3 ) and ( x = 0 ). Checking the intervals:

For ( x < -3 ): both factors are negative, so ( \frac{dy}{dx} > 0
For ( -3 < x < 0 ): the first factor is negative, second is positive, hence ( \frac{dy}{dx} < 0
For ( x > 0 ): both factors are positive, so ( \frac{dy}{dx} > 0

Thus, ( \frac{dy}{dx} < 0 ) in the range:

$(-3, 0)$

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 4 - 2019 - Paper 2

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 4 - 2019 - Paper 2

Show that \( \frac{dy}{dx} = \frac{A}{(x + 1)^{n}} \) where A and n are constants to be found.

Hence deduce the range of values for x for which \( \frac{dy}{dx} < 0 \)

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