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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 use the quotient rule. Let:

$u = \ln(x^{2}+1)\quad v = x$

Thus, we have:

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

Calculating $\frac{du}{dx}$ :

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

Now substituting:

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

Simplifying gives:

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

Step 2

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Answer

Starting with $x = \tan y$ , differentiating both sides with respect to $x$ gives:

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

Now, applying the chain rule:

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

Using the identity $\sec^{2} y = 1 + \tan^{2} y$ , we can express:

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

Substituting back $x = \tan y$ gives us:

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

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