x/2 is the same as (1/2) times x. Thus, you can use the formula for a constant factor.
The derivative of ln x is 1/x The derivative of 2ln x is 2(1/x) = 2/x
m
e^(-2x) * -2 The derivative of e^F(x) is e^F(x) times the derivative of F(x)
Your expression simplifies to just x^2 {with the restriction that x > 0}. The derivative of x^2 is 2*x
sqrt(x) = x^(1/2) The derivative is (1 / 2) * x^(-1 / 2) = 1 / (2 * x^(1 / 2)) = 1 / (2 * sqrt(x))
Following the correct order of operations: derivative of x^2 + 6/2 = derivative of x^2 +3, which equals 2x
(1/2(x^-1/2))/x
2
d/dx((√12)/x)=-(√12)/x^2
Well if you have 5/X then you can rewrite this like 5x-1. And the derivative to that is -5x-2 and that can be rewrote to: -(5/x2).
The derivative of ln x is 1/x The derivative of 2ln x is 2(1/x) = 2/x
13
1 divided by x to the third power equals x to the negative third. The derivative of x to the negative third is minus three x to the negative fourth.
Negative the derivative of f(x), divided by f(x) squared. -f'(x) / f²(x)
-4/x2
Derivative of x = 1, and since sqrt(x) = x^(1/2), derivative of x^(1/2) = (1/2)*(x^(-1/2))Add these two terms together and derivative = 1 + 1/(2*sqrt(x))
So basically this is just a quotient rule problem with the chain rule Turn the square root into the 1/2 power and the derivative of the bottom with the chain rule is 1/2(1+x^2)^-1/2 and add on the derivative of the inside, 2x the full derivative is ((1+x^2)(1)-(x)(1/2(1+x^2)^-1/2 +2x))/ 1+x^2 since you square the denominator when you apply the quotient rule.