A-Level Edexcel Maths Pure Implicit Differentiation: 8. (a) By writing sec θ = \frac{

8. (a) By writing sec θ = \frac{1}{cos θ}, show that \frac{d}{dθ}(sec θ) = sec θ tan θ (b) Given that $ x = e^{sec y} $, $ x > e, \; 0 < y < \frac{π}{2} $ show that $ \frac{dy}{dx} = \frac{1}{x g(x)} $, \; $ x > e $ where g(x) is a function of ln x. - Edexcel - A-Level Maths Pure - Question 8 - 2018 - Paper 5

Question 8

$8.-(a)-By-writing-sec-θ-=-\frac{1}{cos-θ},-show-that-\frac{d}{dθ}(sec-θ)-=-sec-θ-tan-θ---(b)-Given-that-$-x-=-e^{sec-y}-$,-$-x->-e,-\;-0-<-y-<-\frac{π}{2}-$--show-that-$-\frac{dy}{dx}-=-\frac{1}{x-g(x)}-$,-\;-$-x->-e-$--where-g(x)-is-a-function-of-ln-x.-Edexcel-A-Level Maths Pure-Question 8-2018-Paper 5.png$

8. (a) By writing sec θ = \frac{1}{cos θ}, show that \frac{d}{dθ}(sec θ) = sec θ tan θ (b) Given that $ x = e^{sec y} $, $ x > e, \; 0 < y < \frac{π}{2} $ sho... show full transcript

Worked Solution & Example Answer:8. (a) By writing sec θ = \frac{1}{cos θ}, show that \frac{d}{dθ}(sec θ) = sec θ tan θ (b) Given that $ x = e^{sec y} $, $ x > e, \; 0 < y < \frac{π}{2} $ show that $ \frac{dy}{dx} = \frac{1}{x g(x)} $, \; $ x > e $ where g(x) is a function of ln x. - Edexcel - A-Level Maths Pure - Question 8 - 2018 - Paper 5

Step 1

By writing sec θ = \frac{1}{cos θ}, show that \frac{d}{dθ}(sec θ) = sec θ tan θ

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Answer

To find ( \frac{d}{dθ}(sec θ) ), we can express sec θ as ( \frac{1}{cos θ} ).

Utilizing the chain rule:

$\frac{d}{dθ}(sec θ) = \frac{d}{dθ}\left(\frac{1}{cos θ}\right) = -\frac{1}{(cos θ)^2} \cdot \frac{d}{dθ}(cos θ)$

We know ( \frac{d}{dθ}(cos θ) = -sin θ ), so:

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

Thus, we have shown that ( \frac{d}{dθ}(sec θ) = sec θ tan θ ).

Step 2

Given that $ x = e^{sec y} $, \; x > e, \; 0 < y < \frac{π}{2} \) show that $ \frac{dy}{dx} = \frac{1}{x g(x)} $, \; $ x > e $

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Answer

Starting from the equation ( x = e^{sec y} ), we differentiate both sides with respect to x:

( \frac{dx}{dy} = e^{sec y} sec y tan y )

Now we rearrange for ( \frac{dy}{dx} ):

$\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{e^{sec y} sec y tan y}$

Next, we rewrite ( e^{sec y} ) in terms of x:

$= \frac{1}{x sec y tan y}$

Recognizing that the function g(x) relates to ln x, we can establish that:

$g(x) = sec y = e^{\ln x} = x$

Thus, we have confirmed ( \frac{dy}{dx} = \frac{1}{x g(x)} ), where g(x) is a function of ln x.

By writing sec θ = \frac{1}{cos θ}, show that \frac{d}{dθ}(sec θ) = sec θ tan θ

Given that \( x = e^{sec y} \), \; x > e, \; 0 < y < \frac{π}{2} \) show that \( \frac{dy}{dx} = \frac{1}{x g(x)} \), \; \( x > e \)

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