Question

Find the derivative of f(x)=3/sqrt(x^3-2)

Answer 1

\[\Large f(x)={3 \over \sqrt{x^3-2}}\]
correct?

Answer 2

Yes

Answer 3

Alright do you know hot to do the chain rule?

Answer 4

@jamie1522 do you know how?

Answer 5

Yea sort of

Answer 6

dy/dx= vdu/dx-udv/dx all over v2

Answer 7

Ok, well it just means that you have to take the derivative of everything in your "outer function" and multiply it by the derivative of your "inner function"

\[\Large {dy \over dx}f(g(x))=f \prime (g(x)) \cdot g \prime(x)\]

Answer 8

Now f(x)= \[\Large {3 \over \sqrt{x}}\]
and g(x)= \[\Large x^3-2\]

Now I want you to first off, take the derivative of f(x) and g(x) and then we'll go from there

Answer 9

with my approach
u = numerator and v = denominator
for this problem u = 3 and v = (x^3-2)^1/2

Answer 10

continue with @doulikepiecauseidont since he/she responded first.
the answer should be the same

Answer 11

\[f'(x)=-\frac{ 3 }{ 2x ^{\frac{ 3 }{ 2 }} }\]

and

\[g'(x)=3x ^{2}\]

Answer 12

Alright now with the equation for f'(x), plug in g(x) which is x^3-2
then you want to multiply that by g'(x) which is 3x^2

Answer 13

So that is \[\frac{ -9x ^{2} }{ 2(3x ^{\frac{ 7 }{ 2 }}) }\] correct?

Answer 14

Not exactly, you want to plug in 3x^2-2 into f'(x) wherever there is an x so It would be
\[\Large f'(g(x))=-\frac{3}{2(3x^2-2)^\frac{3}{2}}\]

Answer 15

Now this will be mulitplied by our g'(x) which is 3x^2

Answer 16

Wait why is it 3x^2-2? I thought you said plug in g(x)? which is x^3-2

Answer 17

Sorry, got my exponents mixed up, you are correct it would be x^3-2

Answer 18

Alright. So would it be \[\frac{ -9x ^{2} }{ 2(x ^{\frac{ 9 }{ 2 }}2^{\frac{ 5 }{ 2 }} )} \] ?

Answer 19

\[\Large f'(g(x))=-\frac{3}{2(3x^2-2)^\frac{3}{2}} \] would be multiplied by g'(x)

Answer 20

3x^2

Answer 21

\[\Large f'(g(x))=-\frac{9x^2}{2(3x^2-2)^\frac{3}{2}}\]

Answer 22

Is that the final answer?

Answer 23

Yeah

Answer 24

\[\Large {dy \over dx}f(g(x))=f \prime (g(x)) \cdot g \prime(x)\]

Answer 25

You have to be able to identify the outer function (in our case it was the sqrt function)
And the inner function (x^3-2)

Answer 26

outer=f(x)
inner=g(x)

Answer 27

Okay thank you so much for your help!

Answer 28

Yw

Answer 29

@jamie1522 was that the correct answer?
@doulikepiecauseidont can you check

my answer was -9 x^2 / 2(x^3-2)^1/2

Answer 30

Your exponent is wrong, remember that (x^3-2) was already raised to the 1/2 power (that's where the sqrt came from) and then when you apply the power rule (since it's in the denominator and is then like being raised to the -1/2 power) you get the - on the outside and the 2 in the denominatior times the function raised to the -3/2 or 1/(x)^3/2

Answer 31

my exponent cannot be wrong 1/2 - 1 = -1/2
when you differentiate


maybe yours can be simplified or you made a mistake
look again, that's not the only thing you have different