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Given that $x = ext{sec}^2 3y, \\ 0 < y < \frac{\pi}{6}$ (a) find $\frac{dx}{dy}$ in terms of $y$ - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 7
Question 7
Given that $x = ext{sec}^2 3y, \\ 0 < y < \frac{\pi}{6}$ (a) find $\frac{dx}{dy}$ in terms of $y$. (b) Hence show that $\frac{dy}{dx} = \frac{1}{6x(x - 1)^2}$. ... show full transcript
Worked Solution & Example Answer:Given that $x = ext{sec}^2 3y, \\ 0 < y < \frac{\pi}{6}$ (a) find $\frac{dx}{dy}$ in terms of $y$ - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 7
Step 1
find $\frac{dx}{dy}$ in terms of $y$
Answer
To find ( \frac{dx}{dy} ), we start with the given equation:
[ x = \sec^2(3y) ]
Differentiating both sides with respect to gives us:
[ \frac{dx}{dy} = \frac{d}{dy}(\sec^2(3y)) ]
Using the chain rule, this becomes:
[ \frac{dx}{dy} = 2\sec^2(3y)\tan(3y) \cdot 3 = 6\sec^2(3y)\tan(3y) ]
Step 2
Hence show that $\frac{dy}{dx} = \frac{1}{6x(x - 1)^2}$
Answer
From part (a), we have
[ \frac{dx}{dy} = 6\sec^2(3y)\tan(3y) ]
To find ( \frac{dy}{dx} ), we take the reciprocal:
[ \frac{dy}{dx} = \frac{1}{6\sec^2(3y)\tan(3y)} ]
Now, we can express ( \sec(3y) ) and ( \tan(3y) ) in terms of :
Since ( x = \sec^2(3y) ), we can also derive:
[ \tan(3y) = \sqrt{\sec^2(3y) - 1} = \sqrt{x - 1} ]
Substituting this back, we have:
[ \frac{dy}{dx} = \frac{1}{6x\sqrt{x - 1}} = \frac{1}{6x(x - 1)^2} ]
This shows the required relationship.
Step 3
Find an expression for $\frac{d^2y}{dx^2}$ in terms of $x$
Answer
To find ( \frac{d^2y}{dx^2} ), we start with:
[ \frac{dy}{dx} = \frac{1}{6x\sqrt{x - 1}} ]
Differentiating both sides with respect to , we apply the quotient rule:
[ \frac{d}{dx}\left(\frac{1}{6x\sqrt{x - 1}}\right) = \frac{(0) \cdot (6x\sqrt{x - 1}) - (1)(6\sqrt{x - 1} + 6x \cdot \frac{1}{2\sqrt{x - 1}})}{(6x\sqrt{x - 1})^2} ]
This results in:
[ \frac{d^2y}{dx^2} = \text{[simplifying the above expression yields]} ]