Question

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

Answer 1

(3 x^2+2)/4

Answer 2

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:
\[=(3 x^2+2)/(4 (x^3+2 x+1)^{3/4}\]

Answer 3

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

Answer 4

Thanks you

Answer 5

You are welcome

Answer 6

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,\[\frac{3 x^2+2}{4 \left(x^3+2 x+1\right)^{3/4}} \]