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The curve C has equation $$3^{x+1} + xy - y^2 + 5 = 0$$ Show that $ rac{dy}{dx}$ at the point (1, 3) on the curve C can be written in the form $ rac{1}{ ho} ext{ln}( u e^{3})$, where $ ho$ and $ u$ are integers to be found. - Edexcel - A-Level Maths Pure - Question 4 - 2013 - Paper 1
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The curve C has equation $$3^{x+1} + xy - y^2 + 5 = 0$$ Show that $ rac{dy}{dx}$ at the point (1, 3) on the curve C can be written in the form $ rac{1}{ ho} ext{l... show full transcript
Worked Solution & Example Answer:The curve C has equation $$3^{x+1} + xy - y^2 + 5 = 0$$ Show that $ rac{dy}{dx}$ at the point (1, 3) on the curve C can be written in the form $ rac{1}{ ho} ext{ln}( u e^{3})$, where $ ho$ and $ u$ are integers to be found. - Edexcel - A-Level Maths Pure - Question 4 - 2013 - Paper 1
Step 1
Differentiate the equation implicitly
Answer
Start by differentiating the equation implicitly with respect to x. Applying the chain rule gives us:
rac{d}{dx}(3^{x+1}) + rac{d}{dx}(xy) - rac{d}{dx}(y^2) + 0 = 0
For the first term, we have:
rac{d}{dx}(3^{x+1}) = 3^{x+1} ext{ln}(3)
The second term, using the product rule:
rac{d}{dx}(xy) = x rac{dy}{dx} + y
And for the third term:
rac{d}{dx}(y^2) = 2y rac{dy}{dx}
Combining these gives:
3^{x+1} ext{ln}(3) + x rac{dy}{dx} + y - 2y rac{dy}{dx} = 0
Step 2
Evaluate at the point (1, 3)
Answer
Next, substitute the point (1, 3) into the differentiated equation:
3^{1+1} ext{ln}(3) + 1 rac{dy}{dx} + 3 - 2(3) rac{dy}{dx} = 0
This simplifies to:
9 ext{ln}(3) + rac{dy}{dx} + 3 - 6 rac{dy}{dx} = 0
Rearranging gives:
9 ext{ln}(3) + 3 - 5 rac{dy}{dx} = 0
Thus:
5 rac{dy}{dx} = -9 ext{ln}(3) - 3
Step 3
Solve for \(\frac{dy}{dx}\)
Answer
Now, isolate rac{dy}{dx}:
rac{dy}{dx} = - rac{9 ext{ln}(3) + 3}{5}
To write this in the required form, factor out constants:
rac{dy}{dx} = - rac{3}{5} (3 ext{ln}(3) + 1)
This can be expressed as:
ho} ext{ln}( u e^{3}),$$ where $ ho$ and $ u$ are integers to be determined. In this case, $ ho = -5$ and $ u = 3^{3}$.