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Calculus1 9 Online
OpenStudy (anonymous):

suppose that f(4)=3 and f ' (4)=-5. Find g ' (4) where g(x)=√x f(x).

rvc (rvc):

find dg/dx then put in those values

OpenStudy (anonymous):

but how find dg/dx ?

OpenStudy (anonymous):

for this =√x f(x).

rvc (rvc):

yes

OpenStudy (anonymous):

okey how ?!

rvc (rvc):

do u know the product rule?

OpenStudy (anonymous):

I know that but what mean this f(x) in √x f(x). ?

OpenStudy (amistre64):

its just a function ... the value for it is defined in the given statement so just derive g

OpenStudy (amistre64):

the function g, is composed of the product of the function sqrt(x), and the function f(x) apply the product rule ....

OpenStudy (anonymous):

=1/2(x)^-1/2 . f(x) + (x)^1/2 . ?

OpenStudy (amistre64):

what is the derivative of f(x) ??

OpenStudy (anonymous):

=1 ?

OpenStudy (anonymous):

BRB I come back >

OpenStudy (amistre64):

no .. would you agree that the derivative of g(x) ... is g'(x) ?? if so, the what is the derivative of f(x) notate out to be?

OpenStudy (amistre64):

y derives to y' g derives to g' h derives to h' f derives to ____ ???

OpenStudy (anonymous):

f'

OpenStudy (amistre64):

and its a good thing that they define f(4) and also f'(4) for us doesnt it

OpenStudy (anonymous):

so =1/2(x)^-1/2 . f(4) + (x)^1/2 . f'(4) ?

OpenStudy (amistre64):

close .. when x=4, then you might want to evaluate it all for x=4

OpenStudy (amistre64):

g' (x)=1/2(x)^-1/2 . f(x) + (x)^1/2 . f'(x) g' (4)=1/2(4)^-1/2 . f(4) + (4)^1/2 . f'(4)

OpenStudy (anonymous):

like that ? =1/2(4)^-1/2 . f(4) + (4)^1/2 . f'(4) ?

OpenStudy (amistre64):

exactly

OpenStudy (anonymous):

thanks a lot :)

OpenStudy (amistre64):

yep ;)

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