How do you find f'(-5) using the limit definition given $f (x) = 5 cos(x)$ ?

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Answer 1 · 2017-06-06T19:51:24

$f'(-5)=-4.79$

Recall the limit definition of a derivative is:

$f'(x)-=lim_(h->0)$ $((f(x+h)-f(x))/h)$

$f'(x)=5$ $lim_(h->0)$ $((cos(x+h)-cosx)/h)$

$=5$ $lim_(h->0)$ $((cosxcos h-sinxsin h-cosx)/h)$

$=5$ $lim_(h->0)$ $((cosxcos h-cosx-sinxsin h)/h)$

$=5$ $lim_(h->0)$ $((cosx(cos h-1))/h -(sinxsin h)/h)$

$=5(cosx$ $lim_(h->0)$ $((cos h-1)/h) -sinx$ $lim_(h->0)$ $(sin h/h)$ $)$

$lim_(h->0)$ $sin h/h=1$

$lim_(h->0)$ $(cos h-1)/h=0$

$therefore5(cosx$ $lim_(h->0)$ $((cos h-1)/h) -sinx$ $lim_(h->0)$ $(sin h/h)$ $)$

$=-5sinx$

$f'(x)=-5sinx$

$f'(-5)=-5sin(-5)=-4.79$

Note: the derivative of trig functions can only be evaluated as above given that $x$ is in radians.

How do you find f'(-5) using the limit definition given f (x) = 5 cos(x)f (x) = 5 cos(x)?