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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 7 - 2011 - Paper 5

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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 7-2011-Paper 5.png

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 7 - 2011 - Paper 5

Step 1

Differentiate the Equation

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Answer

To find the gradient, we first differentiate both sides of the equation with respect to xx:

rac{d}{dx}( ext{ln } y) = rac{d}{dx}(2 ext{ln } x)

Using the chain rule on the left, we have:

rac{1}{y} rac{dy}{dx} = 2 rac{1}{x}

Step 2

Solve for dy/dx

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Answer

Rearranging gives us:

rac{dy}{dx} = 2y rac{1}{x} = rac{2y}{x}

Step 3

Find y at x = 2

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Answer

Next, we substitute x=2x = 2 into the original equation to find yy:

extlny=2extln(2) ext{ln } y = 2 ext{ln } (2)

This implies:

y=e2extln(2)=22=4y = e^{2 ext{ln } (2)} = 2^2 = 4

Step 4

Calculate dy/dx at x = 2

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Answer

Now, putting x=2x = 2 and y=4y = 4 into the derivative:

rac{dy}{dx} = rac{2 imes 4}{2} = 4

Step 5

Final Answer

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Answer

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

Gradient = 4

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