(a) By writing $\sec \theta = \frac{1}{\cos \theta}$, show that $\frac{d}{d\theta}(\sec \theta) = \sec \theta \tan \theta$. (b) Given that $x = e^{\sec y}$ where $x > e, \; 0 < y < \frac{\pi}{2}$, show that $\frac{dy}{dx} = \frac{1}{x \cdot g(x)}$ where $g(x)$ is a function of $\ln x$.

A-Level Edexcel Maths Pure Exponential & Logarithms: (a) By writing $\sec \theta = \f

(a) By writing $\sec \theta = \frac{1}{\cos \theta}$, show that $\frac{d}{d\theta}(\sec \theta) = \sec \theta \tan \theta$ - Edexcel - A-Level Maths Pure - Question 2 - 2017 - Paper 4

Question 2

$(a)-By-writing-$\sec-\theta-=-\frac{1}{\cos-\theta}$,-show-that-$\frac{d}{d\theta}(\sec-\theta)-=-\sec-\theta-\tan-\theta$-Edexcel-A-Level Maths Pure-Question 2-2017-Paper 4.png$

(a) By writing $\sec \theta = \frac{1}{\cos \theta}$, show that $\frac{d}{d\theta}(\sec \theta) = \sec \theta \tan \theta$. (b) Given that $x = e^{\sec y}$ wher... show full transcript

Worked Solution & Example Answer:(a) By writing $\sec \theta = \frac{1}{\cos \theta}$, show that $\frac{d}{d\theta}(\sec \theta) = \sec \theta \tan \theta$ - Edexcel - A-Level Maths Pure - Question 2 - 2017 - Paper 4

Step 1

Show that $\frac{d}{d\theta}(\sec \theta) = \sec \theta \tan \theta$

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Answer

To begin, we write (\sec \theta) as (\frac{1}{\cos \theta}). Using the chain rule, we find:

$\frac{d}{d\theta}(\sec \theta) = \frac{d}{d\theta}(\frac{1}{\cos \theta}) = -\frac{1}{\cos^2 \theta} \cdot \frac{d}{d\theta}(\cos \theta) = -\frac{1}{\cos^2 \theta} \cdot (-\sin \theta) = \frac{\sin \theta}{\cos^2 \theta}$

Recognizing that (\frac{\sin \theta}{\cos \theta} = \tan \theta) and (\sec \theta = \frac{1}{\cos \theta}), we derive the final result:

$\frac{d}{d\theta}(\sec \theta) = \sec \theta \tan \theta$

Step 2

Show that $\frac{dy}{dx} = \frac{1}{x g(x)}$

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Answer

Given the equation (x = e^{\sec y}), we can derive (y) in terms of (x):

First, take the natural logarithm of both sides:

$\ln x = \sec y$
We know that (\sec y = \frac{1}{\cos y}), which leads to:

$\cos y = \frac{1}{\ln x}$
To find (\frac{dy}{dx}), we differentiate:

$\frac{dy}{dx} = \frac{1}{x \cdot \sec y} \cdot \frac{d}{dy}(\sec y)$ Using the derivative of (\sec y) as derived before, we get:

$\frac{dy}{dx} = \frac{1}{x \cdot \sec y} \cdot (\sec y \tan y) = \frac{\tan y}{x}$
Substitute back (\tan y = \sqrt{\sec^2 y - 1} = \sqrt{\left(\frac{1}{\cos y}\right)^2 - 1} = \frac{\sqrt{1 - \cos^2 y}}{\cos y} = \frac{\sqrt{\left(\frac{1}{\ln x}\right)^2 - 1}}{\frac{1}{\ln x}} = \ln x \cdot g(x)$$ where this forms the relationship required. Thus we rewrite:

$\frac{dy}{dx} = \frac{1}{x g(x)}$ where (g(x)) is defined as a function of (\ln x).

(a) By writing $\sec \theta = \frac{1}{\cos \theta}$, show that \(\frac{d}{d\theta}(\sec \theta) = \sec \theta \tan \theta\) - Edexcel - A-Level Maths Pure - Question 2 - 2017 - Paper 4

Worked Solution & Example Answer:(a) By writing $\sec \theta = \frac{1}{\cos \theta}$, show that \(\frac{d}{d\theta}(\sec \theta) = \sec \theta \tan \theta\) - Edexcel - A-Level Maths Pure - Question 2 - 2017 - Paper 4

Show that \(\frac{d}{d\theta}(\sec \theta) = \sec \theta \tan \theta\)

Show that \(\frac{dy}{dx} = \frac{1}{x g(x)}\)

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