With v= 2sqrt(3x- 4)= 2(3x- 4)^1/2, v'= 3(3x- 4)^-(1/2)= 3(3x- 4)^(.5-1)= 3(3x- 4)^-.5, not "-1.5". That's your basic error.ASidd said:Homework Statement
(9x-8)/ 2(sqrt(3x-4))
Homework Equations
Vu'-Uv'/v^2
i.e. quotient rule
The Attempt at a Solution
I get
V=2(sqrt(3x-4) u=9x-8
V'=3((3x-4)^-1.5) u'=9
When I put this into the equation and solve I get
9x-8(sqrt(3x-4))/6x-8
By the way, with that power in the denominator, so you will need to use the chain rule, anyway, it might be simpler to write the function as (1/2)(9x- 8)(3x-4)^-1/2 and use the product rule rather than the quotient rule.But the answer is 3(9x-16)/4((3x-4)^1.5)
Help Please?
The purpose of finding a derivative is to determine the rate of change of a function at a specific point. It can also be used to find the slope of a curve or to analyze the behavior of a function.
The derivative of a function can be found using various methods such as the power rule, product rule, quotient rule, and chain rule. These methods involve taking the limit of the difference quotient as the change in input approaches zero.
The most commonly used notation for derivatives is the prime notation, where the derivative of a function f(x) is denoted as f'(x). Other notations include Leibniz notation (dy/dx) and Lagrange notation (Df(x)).
Yes, a derivative can be negative. This indicates that the function is decreasing at that point. A positive derivative indicates that the function is increasing at that point, while a derivative of zero indicates a horizontal tangent.
Derivatives and integrals are inverse operations of each other. The derivative of a function gives the rate of change, while the integral of a function gives the total accumulation. This relationship is known as the Fundamental Theorem of Calculus.