Find the gradient of the curve with equation $$ ext{ln} y = 2 ext{ln} x, \, x > 0, \, y > 0$$ at the point on the curve where $x = 2$ - Edexcel - A-Level Maths Pure - Question 7 - 2011 - Paper 5

Next, we substitute $x = 2$ into the derivative equation:

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

From this, we find:

$\frac{dy}{dx} = y$

To find the corresponding $y$ value when $x = 2$, substitute $x = 2$ back into the original equation:

$\text{ln} y = 2 \text{ln} 2$

Taking the exponential of both sides gives:

$y = e^{2 \text{ln} 2} = 2^2 = 4$

Now, substitute $y = 4$ into the equation for the derivative:

$\frac{dy}{dx} = 4$

Thus, the gradient of the curve at the point where $x = 2$ is:

$\frac{dy}{dx} = 4$