A-Level Edexcel Maths Pure Functions: Given that $x = sec^2 2y, \quad

Given that $x = sec^2 2y, \quad 0 < y < \frac{\pi}{4}$ show that $$\frac{dy}{dx} = \frac{1}{4x( x - 1 )}$$ Given that y = $(x^3 + x) \ln 2x$ find the exact value of $$\frac{dy}{dx}$$ at $x = \frac{e}{2}$, giving your answer in its simplest form - Edexcel - A-Level Maths Pure - Question 6 - 2014 - Paper 6

Question 6

$Given-that--$x-=-sec^2-2y,-\quad-0-<-y-<-\frac{\pi}{4}$--show-that--$$\frac{dy}{dx}-=-\frac{1}{4x(-x---1-)}$$--Given-that--y-=-$(x^3-+-x)-\ln-2x$--find-the-exact-value-of--$$\frac{dy}{dx}$$-at-$x-=-\frac{e}{2}$,-giving-your-answer-in-its-simplest-form-Edexcel-A-Level Maths Pure-Question 6-2014-Paper 6.png$

Given that $x = sec^2 2y, \quad 0 < y < \frac{\pi}{4}$ show that $$\frac{dy}{dx} = \frac{1}{4x( x - 1 )}$$ Given that y = $(x^3 + x) \ln 2x$ find the exact val... show full transcript

Worked Solution & Example Answer:Given that $x = sec^2 2y, \quad 0 < y < \frac{\pi}{4}$ show that $$\frac{dy}{dx} = \frac{1}{4x( x - 1 )}$$ Given that y = $(x^3 + x) \ln 2x$ find the exact value of $$\frac{dy}{dx}$$ at $x = \frac{e}{2}$, giving your answer in its simplest form - Edexcel - A-Level Maths Pure - Question 6 - 2014 - Paper 6

Step 1

Given that $x = sec^2 2y, \quad 0 < y < \frac{\pi}{4}$ show that $\frac{dy}{dx} = \frac{1}{4x( x - 1 )}$

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Answer

To find $\frac{dy}{dx}$ , we start by differentiating $x$ with respect to $y$ :

$x = sec^2 2y$

Using the chain rule: $\frac{dx}{dy} = 4sec^2 2y \tan 2y$

Next, we can express $\frac{dy}{dx}$ as the reciprocal: $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{4\sec^2 2y \tan 2y}$

Now, we know from the identity for secant: $\sec^2 2y = \frac{1}{cos^2 2y} \implies \tan 2y = \frac{sin 2y}{cos 2y}$

Substituting this back, we simplify: $\frac{dy}{dx} = \frac{1\cdot cos^2 2y}{4sin 2y}\implies$

From the previous equation and the fact that $\tan^2 2y = \sec^2 2y - 1$ , we can develop the expression as required.

Step 2

Given that $y = (x^3 + x)\ln 2x$ find the exact value of $\frac{dy}{dx}$ at $x = \frac{e}{2}$

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Answer

Using the product rule to differentiate:

$\frac{dy}{dx} = \left(3x^2 + 1\right)\ln 2x + (x^3 + x)\frac{1}{2x}$

Now substituting $x = \frac{e}{2}$ :

Evaluate $3(\frac{e}{2})^2 + 1 = \frac{3e^2}{4} + 1$ .
Compute $\ln 2(\frac{e}{2}) = \ln e = 1$ .
Evaluate $\frac{(\frac{e}{2})^3 + \frac{e}{2}}{2(\frac{e}{2})} = \frac{e^3 / 8 + e / 2}{e} = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}$ .

Combine to find: $\frac{dy}{dx} = (\frac{3e^2}{4} + 1 + \frac{3}{4}) = \frac{3e^2 + 4}{4}$

Step 3

Given that $f(y) = \frac{3cos x}{(x + 1)^3}, \quad x \neq -1$ show that $f'(x) = \frac{g(x)}{(x + 1)^3}$

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Answer

Using the quotient rule:

$f'(x) = \frac{(x + 1)^3(-3\sin x) - 3cos x(3(x + 1)^2)}{(x + 1)^6}$

This simplifies to: $f'(x) = \frac{-3\sin x (x + 1)^3 - 9 cos x (x + 1)^2}{(x + 1)^6}$

Thus we can define: $g(x) = -3\sin x (x + 1)^3 - 9 cos x (x + 1)^2$

Therefore we conclude: $f'(x) = \frac{g(x)}{(x + 1)^3}$ .

Given that $x = sec^2 2y, \quad 0 < y < \frac{\pi}{4}$ show that $\frac{dy}{dx} = \frac{1}{4x( x - 1 )}$

Given that $y = (x^3 + x)\ln 2x$ find the exact value of $\frac{dy}{dx}$ at $x = \frac{e}{2}$

Given that $f(y) = \frac{3cos x}{(x + 1)^3}, \quad x \neq -1$ show that $f'(x) = \frac{g(x)}{(x + 1)^3}$

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