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4. (i) Given that $x = ext{sec}^2 2y$, $0 < y < \frac{\pi}{4}$ show that \[ \frac{dy}{dx} = \frac{1}{4x(y - 1)} \] (4) (ii) Given that $y = (x^2 + x) \ln 2x$ find the exact value of \[ \frac{dy}{dx} \text{ at } x = \frac{e}{2}, \text{ giving your answer in its simplest form.} \] (5) (iii) Given that $f(y) = \frac{3\cos y}{(x + 1)^3}$, $x \neq -1$ show that \[ f'(x) = \frac{g(x)}{(x + 1)^3}, \text{ where } g(x) \text{ is an expression to be found.} \] $x \neq -1$ - Edexcel - A-Level Maths Pure - Question 6 - 2014 - Paper 6
Question 6
![4.-(i)-Given-that----$x-=--ext{sec}^2-2y$,-$0-<-y-<-\frac{\pi}{4}$----show-that----\[----\frac{dy}{dx}-=-\frac{1}{4x(y---1)}----\]----(4)-----(ii)-Given-that----$y-=-(x^2-+-x)-\ln-2x$----find-the-exact-value-of----\[----\frac{dy}{dx}-\text{-at-}-x-=-\frac{e}{2},----\text{-giving-your-answer-in-its-simplest-form.}----\]----(5)-----(iii)-Given-that----$f(y)-=-\frac{3\cos-y}{(x-+-1)^3}$,----$x-\neq--1$----show-that----\[----f'(x)-=-\frac{g(x)}{(x-+-1)^3},----\text{-where-}-g(x)-\text{-is-an-expression-to-be-found.}----\]----$x-\neq--1$-Edexcel-A-Level Maths Pure-Question 6-2014-Paper 6.png](https://cdn.simplestudy.cloud/assets/backend/uploads/question/MathsPure_2014_6_1_QP_4_2014 .jpg)
4. (i) Given that $x = ext{sec}^2 2y$, $0 < y < \frac{\pi}{4}$ show that \[ \frac{dy}{dx} = \frac{1}{4x(y - 1)} \] (4) (ii) Given that $y =... show full transcript
Worked Solution & Example Answer:4. (i) Given that $x = ext{sec}^2 2y$, $0 < y < \frac{\pi}{4}$ show that \[ \frac{dy}{dx} = \frac{1}{4x(y - 1)} \] (4) (ii) Given that $y = (x^2 + x) \ln 2x$ find the exact value of \[ \frac{dy}{dx} \text{ at } x = \frac{e}{2}, \text{ giving your answer in its simplest form.} \] (5) (iii) Given that $f(y) = \frac{3\cos y}{(x + 1)^3}$, $x \neq -1$ show that \[ f'(x) = \frac{g(x)}{(x + 1)^3}, \text{ where } g(x) \text{ is an expression to be found.} \] $x \neq -1$ - Edexcel - A-Level Maths Pure - Question 6 - 2014 - Paper 6
Step 1
Given that $x = \text{sec}^2 2y$, $0 < y < \frac{\pi}{4}$ show that $\frac{dy}{dx} = \frac{1}{4x(y - 1)}$
Answer
To find (\frac{dy}{dx}), we first need to express in terms of . From the relationship (x = \text{sec}^2(2y)), we can differentiate both sides with respect to :
\[
\frac{dx}{dy} = 2\text{sec}^2(2y)\cdot\text{tan}(2y)\text{.}
\]
Next, we find \(\frac{dy}{dx}\) by taking the reciprocal:
\[
\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{2\text{sec}^2(2y)\cdot\text{tan}(2y)}\text{.}
\]
Using the identity \(\text{sec}^2(2y) = 1 + \text{tan}^2(2y)\) and substituting back, we can show that:
\[
\frac{dy}{dx} = \frac{1}{4x(y - 1)}\text{.}
\]
Step 2
Given that $y = (x^2 + x) \ln 2x$ find the exact value of $\frac{dy}{dx}$ at $x = \frac{e}{2}$
Answer
First, we need to differentiate :
\[
\frac{dy}{dx} = (2x + 1) \ln 2x + (x^2 + x) \cdot \frac{1}{x}\text{.}
\]
Now we evaluate at $x = \frac{e}{2}$:
\[
\frac{dy}{dx} \bigg|_{x = \frac{e}{2}} = \left(2 \cdot \frac{e}{2} + 1\right) \ln \left(2 \cdot \frac{e}{2}\right) + \left(\left(\frac{e}{2}\right)^2 + \frac{e}{2}\right) \cdot \frac{1}{\frac{e}{2}}\text{.}
\]
Simplifying this gives the exact value in its simplest form.
Step 3
Given that $f(y) = \frac{3\cos y}{(x + 1)^3}$, show that $f'(x) = \frac{g(x)}{(x + 1)^3}$ where $g(x)$ is an expression to be found
Answer
To find , we apply the quotient rule. Let
\[ g(y) = 3\cos y \text{ and } h(x) = (x + 1)^3\text{.}
\]
Then,
\[ f'(y) = \frac{g'(y)h(y) - g(y)h'(x)}{(h(x))^2}\text{.}
\]
By differentiating, we find:
\[
g'(y) = -3\sin y,\h'(x) = 3(x + 1)^2\text{.}
\]
Now substituting $g(y)$ and $h(x)$:
\[ f'(x) = \frac{-3\sin y (x + 1)^3 - 3\cos y \cdot 3(x + 1)^2}{(x + 1)^6} = \frac{g(x)}{(x + 1)^3}\text{.}
\]
Thus, we have shown the required result.