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In the statement of the attached snapshot it states:
$1/\cos(x)$ is made up of $1/g$ and $\cos()$:
$$f(g) = 1/g$$
How is $1/\cos(x)$ partially made up of $1/g$?
Where does $1/g$ come from?
1 Answer
$\begingroup$Let $g(x) = \cos x$. Then $\frac{1}{\cos x} = \frac{1}{g}$. Then$$\left(\frac{1}{\cos x}\right)' = \left(\frac{1}{g}\right)' = -\frac{1}{g^2}g' = -\frac{1}{\cos ^2x}(-\sin x) = \frac{\sin x}{\cos^2x}$$
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