## anonymous 5 years ago I need help in finding the derivative of F(X)=(1+2x+x^3)^1/4

1. anonymous

(3 x^2+2)/4

2. anonymous

Possible derivation: $d/dx(x^3+2 x+1)^{1/4}$ Use the chain rule, $d/dx(x^3+2 x+1)^{1/4}=(du ^{1/4})/du \times du/dx$ Where: $u = x^3+2 x+1$ and $( du ^{1/4})/du =1/(4u ^{3/4})$ $=(d/dx(x ^{3}+2x+1))/(4 (x^3+2 x+1)^{3/4}$ Differentiate the sum term by term and factor out constants: $=(d/dx(x^3)+2(d/dx(x))+d/dx(1))/(4 (x^3+2 x+1)^{3/4}$ The derivative of 1 is zero: $=(d/dx(x^3)+2 (d/dx(x))+0)/(4 (x^3+2 x+1)^{3/4}$ The derivative of x is 1: $=(d/dx(x^3)+2 1)/(4 (x^3+2 x+1)^{3/4})$ The derivative of x^3 is 3x^2: $=(3 x^2+2)/(4 (x^3+2 x+1)^{3/4}$

3. anonymous

Pasi is correct. My mistake. Posted the derivative of $( (x^3+2x+1)^1 )/4$

4. anonymous

Thanks you

5. anonymous

You are welcome

6. anonymous

I believe someone today, 19May2011, had an issue with the derivation of the derivative of the expression in the problem statement. I was the first with a posting of the proposed derivative of: (1 + 2 x + x^3)^1/4 Mathematica, the program I use for the calculations on this site, interpreted the expression above, even though there was a circumflex accent symbol, indicating exponentiation, as follows:$\frac{1}{4} \left(1+2 x+x^3\right)$ the derivative of which is:$\frac{1}{4} \left(2+3 x^2\right)$Had there been parenthesis flanking the exponent of 1/4 then Mathematica would have presented $\left(1+2 x+x^3\right)^{1/4}$as the problem expression, the same expression Pasi used to find the correct derivative,$\frac{3 x^2+2}{4 \left(x^3+2 x+1\right)^{3/4}}$