Calculus help?

Question

what is the derivative, and how do you take the derivative of the following?

y= 3( 2x^(-1/2) + 1)^(-1/3)

that reads "three times the quantity two-x to the negative one-half plus one, quantity to the negative one-third."

i hope that's how it reads at least....

Answer 1

Start with

f(x)=3
g(x)=(2x^(-1/2)+1)^(-1/3)

We know that f'(x)=0, but we need to expand on the g'(x):

(2x^(-1/2)+1)^(-1/3):

For this funcion, let

f(x)=x^(-1/3)
g(x)=2x^(-1/2)+1

Then f'(x)=(-1/3)x^(-4/3)
and g'(x)=-x^(-3/2)

Because this is a composite function, we need to use the formula

f'(g(x))*g'(x)

So we get

(1/3)x^(-3/2)*
(2x^(-1/2)+1)^(-4/3)

for our g'(x) in our original function. To combine everything, we need to use the product rule, so for

y= 3( 2x^(-1/2) + 1)^(-1/3)

f(x)=3
f'(x)=0
g(x)=(2x^(-1/2)+1)^(-1/3)
g'(x)=(1/3)x^(-3/2)*
(2x^(-1/2)+1)^(-4/3)
(solved from above)

The product rule is the formula:

f'(x)g(x)+f(x)g'(x)

So your final solution will be

x^(-3/2)(2x^(-1/2)+1)^(-4/3)

Answer 2

Use the power rule and chain rule...

y' = -(2x^(-1/2)+ 1)^(-4/3) *( -1(x^(-3/2)))

(I'm not 100% on this one though)

Answer 3

the answer is
1 over x(x^.5 +2)((x^.5 +2)/x^.5)^(1/3)

my TI-89 gave me that answer.. I got lost trying to work it out by hand.. i hope this kind of gives you a guideline

Answer 4

You need to use the function of a function rule. You start by rewriting it as y = 3z^(-1/3) and differentiate that. Then you multiply by dz/dx.