Desperate help with Derivatives! Please EXPLAIN!?

Question

I have to take the derivative twice. I got the 1st part which is this:

x/(x^2+1)^1/2

Now I have to take the derivative of that? How? I know the answer but dont know how to get it.

ALSO

The 1st AND 2nd derivative of this:

√r + 3√r

That one reads:

the square root of r plus the CUBE root of r. not the square root

Okay and this one I have no idea how to do, i did chain rule my answer is always wrong!!

y= (x+2)^8 (x+3)^6


thanks

Here is the answer to the 1st one:

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

Answer 1

This is what I've come up with:

y = x/(x²+1)^½
y = (x)(x²+1)^-½
dy/dx = (x)(-½)(x²+1)^-3/2(2x) + (x²+1)^-½(1)
dy/dx = 2x²(-½)(x²+1)^-3/2 + (x²+1)^-½
dy/dx = -x²(x²+1)^-3/2 + (x²+1)^-½
dy/dx = -x² / (x²+1)^3/2 - 1/(x²+1)^½
dy/dx = 1/(x²+1)^½ - x² / (x²+1)^3/2


y = √r + [3]√r (meaning cube root).
y = r^½ + r^1/3
dy/dr = ½r^-½ + 1/3r^-2/3
dy/dr = 1/ 2r^½ + 1/ 3r^2/3
dy/dr = 1/ 2√r + 1/ 3 [3]√r² (one divided by 3 root 3 of r²).


y= (x+2)^8 (x+3)^6
dy/dx = (x+2)^8(6)(x+3)^5(1) + (x+3)^6(8)(x+2)^7(1)
dy/dx = 6(x+2)^8(x+3)^5 + 8(x+3)^6(x+2)^7

Answer 2

Wikipedia is your friend:
http://en.wikipedia.org/wiki/Table_of_derivatives

This has a few simple rules that you apply over and over again until you are done.

For example, if
f(x) = x/((x^2 + 1)^(1/2)), let
g(x) = x and
h(x) = (x^2 + 1)^(1/2) so f(x) = g(x)/h(x).

Applying the quotient rule, we get:
f'(x) = (g'(x)h(x) - g(x)h'(x))/((h(x))^2)

To compute h'(x), we use the exponent rule:
d(x^n)/dx = (n)(x^(n-1))

which applies whatever the value of n (integer or not, positive or negative). And the chain rule:

d(f(u))/dx = (df(u)/du)(du(x)/dx)

In this case, u(x) = x^2 + 1, f(u) = u^(1/2) so
df(u)/du = (1/2)u^(-1/2)
df/dx = (df(u)/du) (du/dx)

Roots call for the exponent rule: the square root of x is just x^(1/2), the cube root of x is just x^(1/3), etc.

As for y = (x+2)^8 (x+3)^6,
start with the product rule
then the exponent rule for each term
the chain rule comes down to d(x+2)/dx which is 1

Answer 3

try quotient rule
derivative of x/(x^2+1)^1/2 is

[ [(x²+1)^(1/2)] 1 - x {(1/2) (x²+1)^(-1/2) (2x) } ] / (x²+1)

take out (x²+1)^(-1/2)
this gives
(x²+1)^(-1/2) [ (x²+1) - x²] / (x²+1)

or 1/ [(x²+1)^3/2 ]