Math (Calculus)
posted by Amanda .
y= x^(2x+1)
find the derivative of the function using natural log

Log(y) = (2x+1)Log(x)
Differentiate both sides:
y'/y = 2 Log(x) + 2 + 1/x
Multiply by y:
y' = [2 Log(x) + 2 + 1/x] x^(2x+1) 
y=x2x+1
Find the derivative of the expression.
(2x+1)
To find the derivative of 2x, multiply the base (x) by the exponent (1), then subtract 1 from the exponent (11=0). Since the exponent is now 0, x is eliminated from the term.
1
Since 1 does not contain x, the derivative of 1 is 0.
2x+1=2+0
Add 0 to 2 to get 2.
2x+1=2
Using the chain rule, the derivative of x2x+1 is
x2x+1*2
Multiply x2x+1 by 2 to get 2x2x+1.
2x2x+1
The derivative of y with respect to x is 2x2x+1.
The derivative of y with respect to x is 2x2x+1.
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