Find the gradient of the curve with equation
$$ ext{ln } y = 2 ext{ln } x, ext{ } x > 0, y > 0$$
at the point on the curve where $x = 2$ - Edexcel - A-Level Maths Pure - Question 3 - 2011 - Paper 5
Question 3
Find the gradient of the curve with equation
$$ ext{ln } y = 2 ext{ln } x, ext{ } x > 0, y > 0$$
at the point on the curve where $x = 2$. Give your answer as an ... show full transcript
Worked Solution & Example Answer:Find the gradient of the curve with equation
$$ ext{ln } y = 2 ext{ln } x, ext{ } x > 0, y > 0$$
at the point on the curve where $x = 2$ - Edexcel - A-Level Maths Pure - Question 3 - 2011 - Paper 5
Step 1
Find the derivative
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Answer
To find the gradient of the curve, we first differentiate the equation. Starting with the equation:
extlny=2extlnx
We differentiate both sides with respect to x:
rac{1}{y} rac{dy}{dx} = 2 rac{1}{x}
Multiplying through by y gives:
rac{dy}{dx} = 2y rac{1}{x}
Step 2
Evaluate at $x = 2$
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Answer
Now, we need to evaluate this expression at x=2. First, we find y when x=2:
y = e^{2 ext{ln } 2} = (e^{ ext{ln } 2})^2 = 2^2 = 4$$
So, at $x = 2$, $y = 4$.
Now, substituting $y$ and $x$ back into our derivative:
$$ rac{dy}{dx} = 2(4) rac{1}{2} = 4$$
Step 3
Final answer
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Answer
Thus, the gradient of the curve at the point where x=2 is:
