Applying the Chain Rule to Derivatives with Square Roots

TMNT
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Chain Rule

Question is
Find the derivative of F(x)= 3 sq rt of x^3-1

First step I did was changing the Sq RT to (x^3-1)^3/2
Then I solved it by 3/2(X^3-1)^1/2*3X^2

Another problem very similar
F(X)= 3 SQ RT of X^4+3x+2

Step 1 (X^4+3x+2)^3/2
Then 3/2(X^4+3x+2)*4x^3+3

I know how to do the derivatives my only concern is that 3 in front of the square roots are throwing me off, I just want to know if I'm doing it right.
 
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so, your F(x) = 3*sqrt(x^3-1) ?

if so, F'(x) = 3*[1/2(x^3-1)^-1/2]*3x^2 = (9x^2)/(2sqrt(x^3-1))
 
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It looks like you're trying to put the coefficient out front into the exponent, like saying 3*(x^1/2) = x^3/2, which it is not. With derivatives that coefficient just kinda stays put...
 
3\sqrt{x^3-1} = 3(x^3-1)^{1/2}
\frac{d}{dx}[3(x^3-1)^{1/2}] = \frac{1}{2} 3 (x^3-1)^{-1/2} 3x^2 = \frac{9x^2}{2\sqrt{x^3-1}}

just like jth01 said: 3x^{1/2} does NOT equal x^{3/2}
 
Thanks guys.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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