Ask your own question, for FREE!
Mathematics 7 Online
OpenStudy (anonymous):

Differentiate g(x)=3(sqrt x e^x)

OpenStudy (anonymous):

\[ g(x)=3\sqrt{}xe ^{x}\]

OpenStudy (anonymous):

Please show me step by step thanks

OpenStudy (anonymous):

3(sqrt(x)e^x)

OpenStudy (anonymous):

?

OpenStudy (anonymous):

3root(x)e^x + (3/2)(e^x)/root(x)

OpenStudy (anonymous):

yea

OpenStudy (anonymous):

chain rule first treat rootx as constant and diff e^x and then keep e^x constant and diff rootx

OpenStudy (anonymous):

3root(x)e^x + (3/2)(e^x)/root(x)

OpenStudy (anonymous):

Or just use product rule?

OpenStudy (anonymous):

i didnt learn the chain rule yet

OpenStudy (anonymous):

can you show me please

OpenStudy (anonymous):

oh sorry product rule..i said chain rule by mistake

OpenStudy (anonymous):

Product Rule

OpenStudy (anonymous):

Horatio has a pack of candies containing 4 strawberry, 2 lemon, 1 lime, and 3 orange flavored chews, in random order. If he eats each candy immediately after taking it out of the pack, which multiplication expression can be used to determine the probability that the first two candies he eats are lemon?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

can you show me how to use the poduct rule for this problem

OpenStudy (anonymous):

see if there is a product of 2 functions f(x)g(x) the derivative is f(x)g'(x) + f'(x)g(x)

OpenStudy (anonymous):

so now treat e^x as f(x) and rootx as g(x)

OpenStudy (anonymous):

ok..but where is 3

OpenStudy (anonymous):

3 u put outside..it is constant

OpenStudy (anonymous):

o ok..so it would be like this\[3\sqrt{x}e^x = 3(1/2(x))^-1/2\]

OpenStudy (anonymous):

no sure it thats right

OpenStudy (anonymous):

wheres the e^x gne?

OpenStudy (anonymous):

o well that becomes \[e\]

OpenStudy (anonymous):

and its a constant so i removed it

OpenStudy (anonymous):

\[3\sqrt{x}e ^{x} +3e^{x}/2\sqrt{x}\]

OpenStudy (anonymous):

i dont understand

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!