Differentiate:
y=ln(x/(2x+3))^(1/2)
y=(1/2)ln(x/(2x+3))
y'=0+(1/2)((2x+3-2x)/(2x+3)²)
What's next after the quotient rule?
You don't have to use the quotient rule of differentiation.
y = lnx - (1/2)ln(2x+3)
y' = 1/x -(1/2)[1/(2x+3)]*2
= (1/x) - [1/(2x+3)]
which can be rewitten with a common denominator, if you wish.
Thanks for your help!
After applying the quotient rule for differentiation, the next step would be to simplify the expression and further differentiate it if necessary.
The quotient rule states that to differentiate a function in the form of f(x) = u(x)/v(x), where u(x) and v(x) are differentiable functions, the derivative can be found using the following formula:
f'(x) = (v(x) * u'(x) - u(x) * v'(x)) / (v(x))^2
In the given example, the derivative of y = (1/2)ln(x/(2x+3)) would be:
y' = (2x + 3 - 2x) / (2x + 3)^2
Simplifying the expression, we get:
y' = (3) / (2x + 3)^2
At this point, if you want to simplify further, you can leave it as is since it is already in its simplest form. If you need to differentiate further, you can continue by applying other differentiation rules or methods based on the problem or the context.