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Given that $y = \frac{\ln(x^2 + 1)}{x}$, find $\frac{dy}{dx}$ - Edexcel - A-Level Maths Pure - Question 5 - 2010 - Paper 2

Question 5

$Given-that-$y-=-\frac{\ln(x^2-+-1)}{x}$,-find-$\frac{dy}{dx}$-Edexcel-A-Level Maths Pure-Question 5-2010-Paper 2.png$

Given that $y = \frac{\ln(x^2 + 1)}{x}$, find $\frac{dy}{dx}$. Given that $x = \tan(y)$, show that $\frac{dy}{dx} = \frac{1}{1 + x^2}$.

Worked Solution & Example Answer:Given that $y = \frac{\ln(x^2 + 1)}{x}$, find $\frac{dy}{dx}$ - Edexcel - A-Level Maths Pure - Question 5 - 2010 - Paper 2

Step 1

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Answer

To find $\frac{dy}{dx}$ , we will use the quotient rule.

Let ( u = \ln(x^2 + 1) ), then using the product rule, we find:

$\frac{du}{dx} = \frac{2x}{x^2 + 1}$
Applying the quotient rule:

$\frac{dy}{dx} = \frac{ v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$
where ( v = x ) and ( \frac{dv}{dx} = 1 ):

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

Thus,

$\frac{dy}{dx} = \frac{\frac{2x^2}{x^2 + 1} - \ln(x^2 + 1)}{x^2}.$

This gives us the required derivative.

Step 2

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Answer

Given that ( x = \tan(y) ), we differentiate both sides with respect to $x$ :

Using implicit differentiation, we have:

$\frac{dx}{dy} = \sec^2(y)$

So,

$\frac{dy}{dx} = \frac{1}{\sec^2(y)}.$
Employing the identity ( \sec^2(y) = 1 + \tan^2(y) ):

$\frac{dy}{dx} = \frac{1}{1 + \tan^2(y)}.$
Since ( x = \tan(y) ), we substitute:

$\frac{dy}{dx} = \frac{1}{1 + x^2},$ as required.

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