How do you differentiate $f(x) =2xcos xtanx$ ?

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Answer 1 · 2017-01-22T16:03:02

$f'(x) = 2(sinx + xcosx)$

Rewrite $tanx$ as $sinx/cosx$ :

$f(x) = 2xcosx(sinx/cosx)$

$f(x) = 2xsinx$

Differentiate using the product rule. Let $f(x) = g(x) * h(x)$ , with $g(x) = 2x$ and $h(x) =sinx$ . Then $g'(x) = 2$ and $h'(x) = cosx$ .

The product rule states that $f'(x) = g'(x)h(x) + h'(x)g(x)$ .

$f'(x) = 2sinx + 2xcosx$

$f'(x) = 2(sinx + xcosx)$

Hopefully this helps!

How do you differentiate f(x) =2xcos xtanx f(x) =2xcos xtanx ?