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 (1-1=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.
To find the derivative of the function y = x^(2x+1) using natural logarithm, you can take the natural logarithm of both sides of the equation and then differentiate implicitly.
Step 1: Take the natural logarithm of both sides of the equation:
ln(y) = ln(x^(2x+1))
Step 2: Use the logarithmic rule, which states that ln(a^b) = b * ln(a), to simplify the equation:
ln(y) = (2x + 1) * ln(x)
Step 3: Differentiate both sides of the equation implicitly with respect to x. Keep in mind that ln(y) is a composite function, so you will need to use the chain rule.
d/dx[ln(y)] = d/dx[(2x + 1) * ln(x)]
Step 4: Apply the chain rule to differentiate the left side of the equation:
1/y * dy/dx = (2x + 1) * d/dx[ln(x)] + ln(x) * d/dx[2x + 1]
Step 5: Simplify the derivatives on the right side of the equation:
1/y * dy/dx = (2x + 1) * 1/x + ln(x) * 2
Step 6: Multiply both sides of the equation by y:
dy/dx = y * [(2x + 1)/x + 2ln(x)]
Step 7: Substitute the original function y = x^(2x+1) back into the equation:
dy/dx = x^(2x+1) * [(2x + 1)/x + 2ln(x)]
So, the derivative of the function y = x^(2x+1) using natural logarithm is dy/dx = x^(2x+1) * [(2x + 1)/x + 2ln(x)].