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# Tangent of f(x)

If $$f(\cos(2x)) = \sec^2 (x) - 2\sin^2 (x) + 1$$, find the rate of change of $$y=f(x)$$ when $$x =\frac12$$.

Note by Majed Musleh
2 years, 3 months ago

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Using 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 $$f'(\frac{1}{2}) = -\dfrac{2}{\frac{9}{4}} + 1 = \dfrac{1}{9}.$$ · 2 years, 3 months ago