How do you differentiate $g(x) = (cotx + cscx)(tanx - sinx)$ ?

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Answer 1 · 2015-12-06T16:14:11

$g'(x)=secxtanx+sinx$

First, simplify $g(x)$ .

$g(x)=cotxtanx-cotxsinx+cscxtanx-cscxsinx$

$g(x)=1-cosx+secx-1$

$g(x)=secx-cosx$

Now, all we need to find $g'(x)$ are the following identities:

$d/dx[secx]=secxtanx$

$d/dx[cosx]=-sinx$

Thus, $g'(x)=secxtanx+sinx$

This can also be written as: $g'(x)=sinx(sec^2x+1)$

How do you differentiate g(x) = (cotx + cscx)(tanx - sinx) g(x) = (cotx + cscx)(tanx - sinx) ?