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 $x =\frac12$ .

#Calculus

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Comments

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 $f'(\frac{1}{2}) = -\dfrac{2}{\frac{9}{4}} + 1 = \dfrac{1}{9}.$

Brian Charlesworth - 6 years ago

@Brian Charlesworth @Pi Han Goh

Majed Musleh - 6 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$