Determine the derivative of the function:
Y=1 / (Sq.root of (3x-1)
I know the answer is -3 / (2(Sq.root of 3x-1)^3
I just need some aid on the steps.
Thanks in advance.
What if we wrote it like this ...
y = 1/√(3x-1)
= (3x-1)^(-1/2)
now we can use the chain rule
dy/dx = (-1/2)(3x-1)^(-3/2)(3)
= (-3/2)/(3x-1)^(3/2)
= -3/[2√(3x-1)^3
dy/dx=3/2(3x-1)61/2
dy/dx=3/2(3x-1)^1/2
My bad.... apparently didn't hit the shift key for the ^ sign...
To find the derivative of the function Y = 1 / sqrt(3x-1), we can use the quotient rule of differentiation.
The quotient rule states that if we have a function of the form f(x) = g(x) / h(x), then its derivative is given by:
[f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2]
Let's apply this rule to our function:
Y = 1 / sqrt(3x-1)
First, we need to find the derivative of the numerator and the denominator separately.
The derivative of the numerator (g(x)) which is 1, will be zero since it's a constant.
Now, let's find the derivative of the denominator (h(x)):
h(x) = sqrt(3x-1)
To do this, we can use the chain rule.
Let u = 3x-1. Then, h(x) = sqrt(u).
Now, we take the derivative of h(x) with respect to u:
h'(x) = (1/2) * (u)^(-1/2) * du/dx
To find du/dx, we differentiate u = 3x-1 with respect to x:
du/dx = 3
Substituting back into h'(x), we have:
h'(x) = (1/2) * (u)^(-1/2) * 3
h'(x) = 3/2 * (3x-1)^(-1/2)
Now, let's plug these values into the quotient rule equation:
Y' = [(0 * sqrt(3x-1) - 1 * (3/2)*(3x-1)^(-1/2)) / (sqrt(3x-1))^2]
Simplifying further:
Y' = (-3/2)*(3x-1)^(-1/2) / (3x-1)
Y' = -3 / (2*(3x-1)^(3/2))
Thus, the derivative of the function Y = 1 / sqrt(3x-1) is -3 / (2*(sqrt(3x-1))^3).