How do you find f'(0) given $f (x) = {(\frac{x^{5} + 5}{3 cos x})}^{2}$ $f(x) =( (x^5+5)/(3cosx))^2$ ?

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Answer 1 · 2016-12-22T00:38:08

$f'(0) = 2((0^5+5)/(3cos0)) *0 = 0$ See below.

$f'(x) = 2((x^5+5)/(3cosx)) * d/dx((x^5+5)/(3cosx))$

Notice that when we apply the quotient rule to find

$d/dx((x^5+5)/(3cosx))$ ,

The derivatives of the numerator ( $5x^4$ ) and denominator $(-3sinx)$ are both $0$ at $x=0$ , so lets not bother with it.

$f'(0) = 2((0^5+5)/(3cos0)) *0 = 0$

How do you find f'(0) given f(x) =( (x^5+5)/(3cosx))^2f(x)=(x5+53cosx)2f(x) =( (x^5+5)/(3cosx))^2?