hsd said:Homework Statement
F(3)=−2, g(3)=9, f′(3)=−2, and g′(3)=2, find the following numbers:
(a) (f+g)′(3)
(b) (fg)′(3)
(c) (f/g)′(3)
(d) (f/(f−g))′(3)
The Attempt at a Solution
I already have (a) and (b) [a=0 and b=-22]
for (c) i tried:
(g(x)*f'(x) - f(x)g'(x)) / (g(x))^2
evaluate at 3:
(9)(-2) - (-2)(2) / 4
-18 + 4 / 4
-14/4
-7/2
WHICH IS THE WRONG ANSWER
(d) I just can't get started.
Yes, use the quotient rule. After you get the derivative, evaluate the derivative at x = 3.berkeman said:And for d), can you still just use the quotient rule? (I'm not sure, but try it...)
Mark44 said:Yes, use the quotient rule. After you get the derivative, evaluate the derivative at x = 3.
You have three mistakes above:hsd said:I think I am applying the rule wrong. This is what I did:
(f-g)'(f)-(f)'(f-g)/(f-g)
It would be helpful if you included the arguments for the general derivative.hsd said:[(-2)(-2)]'[-2]-[-2]'[(-2)(-9)]/[(-2)(-9)]
8-22/-11
-14/-11 (wrong answer)
Can you please tell me what it is that I am doing wrong?
A derivative in calculus is a mathematical concept that represents the rate of change of a function with respect to its variable. In simpler terms, it measures how much a function changes when its input value changes.
To find a derivative in calculus, you use the process of differentiation. This involves applying specific rules and formulas to a given function to find its derivative. The most common method is using the power rule, where you multiply the coefficient of the variable by its exponent and decrease the exponent by 1.
The purpose of finding derivatives in calculus is to understand the behavior of a function, particularly how it changes over time. Derivatives are used to calculate rates of change, find maximum and minimum values, and solve optimization problems in various fields such as physics, economics, and engineering.
To find the derivative of a function at a specific point, you use the process of differentiation and substitute the given value for the variable in the resulting derivative. This will give you the slope of the tangent line at that point, which represents the rate of change of the function at that point.
Yes, you can find multiple derivatives of a function by repeating the process of differentiation. The first derivative gives you the instantaneous rate of change, while the second derivative gives you the rate of change of the first derivative. The third derivative represents the rate of change of the second derivative, and so on.