Letting \(v = \cos(2x)\) we can write the function as \(f(v) = \dfrac{2}{1 + v} + v,\) and since \(v\) can now just be considered a "dummy" variable we can re-label it as \(x\) to end up with the function in the form \(f(x) = \dfrac{2}{1 + x} + x.\)

Now \(f'(x) = -\dfrac{2}{(1 + x)^{2}} + 1,\) and so \(f'(\frac{1}{2}) = -\dfrac{2}{\frac{9}{4}} + 1 = \dfrac{1}{9}.\)
–
Brian Charlesworth
·
2 years, 1 month ago

## Comments

Sort by:

TopNewestUsing the identity \(\cos(2x) = 2\cos^{2}(x) - 1 \Longrightarrow \cos^{2}(x) = \dfrac{1}{2}(1 + \cos(2x))\) we have that

\(f(\cos(2x)) = \dfrac{1}{\cos^{2}(x)} - 2(1 - \cos^{2}(x)) + 1 = \dfrac{2}{1 + \cos(2x)} - 2 + (1 + \cos(2x)) + 1 = \dfrac{2}{1 + \cos(2x)} + \cos(2x).\)

Letting \(v = \cos(2x)\) we can write the function as \(f(v) = \dfrac{2}{1 + v} + v,\) and since \(v\) can now just be considered a "dummy" variable we can re-label it as \(x\) to end up with the function in the form \(f(x) = \dfrac{2}{1 + x} + x.\)

Now \(f'(x) = -\dfrac{2}{(1 + x)^{2}} + 1,\) and so \(f'(\frac{1}{2}) = -\dfrac{2}{\frac{9}{4}} + 1 = \dfrac{1}{9}.\) – Brian Charlesworth · 2 years, 1 month ago

Log in to reply

@Brian Charlesworth @Pi Han Goh – Majed Musleh · 2 years, 1 month ago

Log in to reply