A-Level Edexcel Maths Pure Exponential & Logarithms: Given the function: $y =

Given the function: $y = \frac{4x}{x^2 + 5}$ (a) Find $\frac{dy}{dx}$, writing your answer as a single fraction in its simplest form - Edexcel - A-Level Maths Pure - Question 4 - 2016 - Paper 3

Question 4

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Given the function: $y = \frac{4x}{x^2 + 5}$ (a) Find $\frac{dy}{dx}$, writing your answer as a single fraction in its simplest form. (b) Hence find th... show full transcript

Worked Solution & Example Answer:Given the function: $y = \frac{4x}{x^2 + 5}$ (a) Find $\frac{dy}{dx}$, writing your answer as a single fraction in its simplest form - Edexcel - A-Level Maths Pure - Question 4 - 2016 - Paper 3

Step 1

Find $\frac{dy}{dx}$

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Answer

To find the derivative $\frac{dy}{dx}$ of the function $y = \frac{4x}{x^2 + 5}$ , we will use the quotient rule. The quotient rule states that if we have a function of the form $\frac{u}{v}$ , then the derivative is given by:

$\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$

Here, let:

$u = 4x$ , giving us $\frac{du}{dx} = 4$ .
$v = x^2 + 5$ , giving us $\frac{dv}{dx} = 2x$ .

Now substituting into the quotient rule:

$\frac{dy}{dx} = \frac{(x^2 + 5)(4) - (4x)(2x)}{(x^2 + 5)^2}$

Simplifying this:

Expanding the numerator: $4x^2 + 20 - 8x^2$ $= -4x^2 + 20$
Therefore, the derivative becomes: $\frac{dy}{dx} = \frac{-4x^2 + 20}{(x^2 + 5)^2}$
Finally, we can write this as: $\frac{dy}{dx} = \frac{20 - 4x^2}{(x^2 + 5)^2}$

So, the answer in simplest form is:

$\frac{20 - 4x^2}{(x^2 + 5)^2}$

Step 2

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

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Answer

To find the values of $x$ for which $\frac{dy}{dx} < 0$ , we need to analyze the expression:

$\frac{20 - 4x^2}{(x^2 + 5)^2} < 0$

The denominator $(x^2 + 5)^2$ is always positive since $x^2 + 5 > 0$ for all real $x$ .
Hence, we focus on the numerator: $20 - 4x^2 < 0$ This simplifies to: $20 < 4x^2$ Dividing both sides by 4 gives: $5 < x^2$ or equivalently, $x^2 > 5$
Taking square roots yields: $|x| > \sqrt{5}$ Therefore, the set of values of $x$ for which $\frac{dy}{dx} < 0$ is:

$x < -\sqrt{5} \quad \text{or} \quad x > \sqrt{5}$

Given the function: $y = \frac{4x}{x^2 + 5}$ (a) Find $\frac{dy}{dx}$, writing your answer as a single fraction in its simplest form - Edexcel - A-Level Maths Pure - Question 4 - 2016 - Paper 3

Worked Solution & Example Answer:Given the function: $y = \frac{4x}{x^2 + 5}$ (a) Find $\frac{dy}{dx}$, writing your answer as a single fraction in its simplest form - Edexcel - A-Level Maths Pure - Question 4 - 2016 - Paper 3

Find $\frac{dy}{dx}$

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

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