Find the derivative of
y = 3x+2/x+1
y=3+2ln|x|
y = 3x + 2(x+1)^-1
dy/dx = 3 -2(x+1)^-2 or
= 3 - 2/(x+1)^2
y = (3x+2)/(x+1)
y' = [(3)(x+1) - (3x+2)(1)]/(x+1)^2
= (3x + 3 - 3x - 2)/(x+1)^2
= 1/(x+1)^2
See what a difference proper punctuation makes? Three reasonable interpretations of your problem, all giving different solutions.
USE PARENTHESES FOR CLARITY!
To find the derivative of the function y = 3x+2/x+1, we can use the quotient rule. The quotient rule states that if we have a function in the form u/v, where u and v are both functions of x, then the derivative of the function is given by:
(dy/dx) = (u'v - uv') / v^2
In this case, u = 3x + 2 and v = x + 1. Now, let's find the derivatives of u and v separately.
To find the derivative of u = 3x + 2, we can use the power rule, which states that if we have a function in the form ax^n, then the derivative is given by:
(du/dx) = n * ax^(n-1)
The derivative of u = 3x + 2 is therefore:
(du/dx) = 3
To find the derivative of v = x + 1, we again use the power rule. In this case, n = 1 and a = 1:
(dv/dx) = 1 * 1^(1-1) = 1
Now, let's find the derivatives of u and v separately. From the quotient rule, we have:
(dy/dx) = [(u'v - uv') / v^2]
Substituting in the values of u', v', u, and v, we get:
(dy/dx) = (3(x + 1) - (3x + 2)(1)) / (x + 1)^2
= (3x + 3 - 3x - 2) / (x + 1)^2
= 1 / (x + 1)^2
Therefore, the derivative of y = 3x + 2/x + 1 is dy/dx = 1 / (x + 1)^2.