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Given that $y = \frac{\ln(x^2+1)}{x}$, find $\frac{dy}{dx}$ - Edexcel - A-Level Maths Pure - Question 4 - 2010 - Paper 2

Question 4

$Given-that-$y-=-\frac{\ln(x^2+1)}{x}$,-find-$\frac{dy}{dx}$-Edexcel-A-Level Maths Pure-Question 4-2010-Paper 2.png$

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 4 - 2010 - Paper 2

Step 1

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

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Answer

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

$u = \ln(x^2 + 1)$

Now, we have:

$\frac{du}{dx} = \frac{2x}{x^2 + 1}$

Using the quotient rule:

$\frac{dy}{dx} = \frac{u'v - uv'}{v^2}$

where $u = \ln(x^2 + 1)$ and $v = x$ . Then:

$u' = \frac{2x}{x^2 + 1}$
$v' = 1$

Substituting in:

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

Simplifying this further yields:

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

Step 2

Given that $x = \tan y$, show that $\frac{dy}{dx} = \frac{1}{1+x^2}$

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Answer

We start with the relationship $x = \tan y$ , which leads us to differentiate:

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

Then, the relation:

$\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{\sec^2 y}$

This can also be expressed as:

$\frac{dy}{dx} = \frac{\cos^2 y}{1}$

Recall the identity for tangent:

$\tan^2 y + 1 = \sec^2 y$

Thus:

$\sec^2 y = 1 + \tan^2 y = 1 + x^2$

Therefore, we have:

$\frac{dy}{dx} = \frac{1}{1 + x^2}$ , which completes the proof.

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