A-Level Edexcel Maths Pure Graphs of Functions: Given that $y = \frac{\ln(x^2 +

Given that $y = \frac{\ln(x^2 + 1)}{x}$, find $ \frac{dy}{dx} $ - Edexcel - A-Level Maths Pure - Question 6 - 2010 - Paper 2

Question 6

$Given-that-$y-=-\frac{\ln(x^2-+-1)}{x}$,-find-$-\frac{dy}{dx}-$-Edexcel-A-Level Maths Pure-Question 6-2010-Paper 2.png$

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 6 - 2010 - Paper 2

Step 1

96%

114 rated

Answer

To find ( \frac{dy}{dx} ), we will apply the quotient rule, which states:

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

where:

Now, we calculate the derivatives:

Compute ( \frac{du}{dx} ):
- Using the chain rule: $\frac{du}{dx} = \frac{1}{x^2+1} \cdot \frac{d}{dx}(x^2 + 1) = \frac{2x}{x^2 + 1}$
Compute ( \frac{dv}{dx} ):
- This gives ( \frac{dv}{dx} = 1 ).

Substituting these into the quotient rule:

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

Thus, combining the terms provides:

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

Step 2

99%

104 rated

Answer

Since we know that ( x = \tan(y) ), we will differentiate both sides with respect to ( x ).

We can use implicit differentiation:

Differentiate both sides:
- This leads to: $\frac{dx}{dy} = \sec^2(y)$
Inverting this gives: $\frac{dy}{dx} = \frac{1}{\sec^2(y)}$
Using the identity ( \sec^2(y) = 1 + \tan^2(y) ), we substitute back:
- Since ( x = \tan(y) ), we have: $\sec^2(y) = 1 + x^2$

Thus:

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

This shows the required result.