SOLUTION: i need the derivative of {{{ x+3*x^(2/3) }}} i've done it on paper so many times but it doesn't seem to come out right. there must be some way for the derivative to either

Algebra ->  Exponents-negative-and-fractional -> SOLUTION: i need the derivative of {{{ x+3*x^(2/3) }}} i've done it on paper so many times but it doesn't seem to come out right. there must be some way for the derivative to either       Log On


   

Question 248504: i need the derivative of
+x%2B3%2Ax%5E%282%2F3%29+
i've done it on paper so many times but it doesn't seem to come out right.
there must be some way for the derivative to either equal 0 or be undefined, and none have my answers have allowed for this. please help!
then i also need the second derivative as well..
Found 2 solutions by stanbon, jsmallt9:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
i need the derivative of
y = x+3*x^(2/3)
--------------------
dy/dx = 1 + 3(2/3)x^(-1/3)
= 1 + 2x^(-1/3)
========================
Cheers,
Stan H.

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29+=+x+%2B+3x%5E%282%2F3%29
The derivative of a sum is the sum of the derivatives. So
f'(x) = %28d%2F%28dx%29%29%28x%29+%2B+%28d%2F%28dx%29%293x%5E%282%2F3%29
I hope you know that the derivative of x is 1. For the derivative of 3x%5E%282%2F3%29 we will use the bring the exponent down in front and subtract 1. This gives us:
f'(x) = 1+%2B+%282%2F3%293x%5E%28%282%2F3%29-1%29
which simplifies to:
f'(x) = 1+%2B+2x%5E%28%28-1%29%2F3%29

For the second derivative we find the derivative of the first derviative. The derivative of a constant like 1 is zero. And we will repeat the exponent thing on the second term:
f''(x) = 0+%2B+%28%28-1%29%2F3%29x%5E%28%28%28-1%29%2F3%29-1%29
which simplifies to:
f''(x) = %28%28-1%29%2F3%29x%5E%28%28-4%29%2F3%29

These are the first and second derivatives. You mention something about them being zero. As a general rule derivative are not always zero.

However we are often interested in the value(s) of x that make the derivatives zero. If this is what you are looking for then set the derivatives equal to zero and use Algebra to solve for x. Here's a solution for the x values that make the first derivative zero:
0+=+1+%2B+2x%5E%28%28-1%29%2F3%29
Subtract 1 from each side:
-1+=+2x%5E%28%28-1%29%2F3%29
Divide by 2:
%28-1%29%2F2+=+x%5E%28%28-1%29%2F3%29
Raise each side to the -3 power:
%28%28-1%29%2F2%29%5E%28-3%29+=+%28x%5E%28%28-1%29%2F3%29%29%5E%28-3%29
-8+=+x
This tells us that the slope of the tangent to f(x) is zero at x = -8.

I will leave it up to you to find the x values, if any, that make the second derivative zero.