Calculate Derivatives: Using Information to Find (fg)'(1) and (fog)'(1)

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In summary, we are given information about two functions, f and g, and are asked to use it to calculate two derivatives: (fg)'(1) and (f o g)'(1). We start by using the product rule to calculate (fg)'(x) and then plug in the given values to find (fg)'(1). For the second part, we use the chain rule to calculate (f o g)'(x) and then again plug in the given values to find (f o g)'(1).
  • #1
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Homework Statement


Suppose f and g are functions. Assume that f is differentiable and f(1) = 2, f'(1)=4 and that g(x) = x^4 -x+1. Use this information to calculate:

(fg)'(1)
(f o g)'(1)

Homework Equations


The Attempt at a Solution



I'm unsure of what this wants me to do Does it want me to multiply the derivatives of f and g together and calculate it at 1 or does it want me to multiply f and g then find the derivative and plug in 1.

For the 2nd one I found the derivative of g(x) to be 4x^3 -1 plugged a 1 into that and got 3 now i have f'(3) = ? how can i know that when they only gave me f(1) and f'(1) ?
 
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  • #2
I would guess that (fg)(x) means f(x) times g(x), but you should check your book or course notes or whatever is relevant for the definition you are supposed to be using.

For the second part, (f o g)(x) usually means f(g(x)), which can be differentiated by the chain rule. You should find that you do not need the value of f'(3).
 
  • #3
Does "product rule" and "chain rule" ring a bell?
Product rule: [tex](fg)'=f'g+fg'[/tex]
and
Chain rule: [tex] (f \circ g)'(x) = f'(g(x)) g'(x)[/tex]

Just a clarification of notation: [tex](f \circ g)(x)[/tex] is also written as [tex]f(g(x))[/tex] and is read as "[tex]f[/tex] of [tex]g[/tex] of [tex]x[/tex]". It simply means you first apply the [tex]g[/tex] function to [tex]x[/tex] and then the [tex]f[/tex] function to the resultant. The [tex]'[/tex] indicates the first derivitive of the function.
 
  • #4
ah ok its not (fg)x tho its (fg)'(x)

and I don't know the equation for f(x) it doesn't give it to you so how can i find the derivative of

f(x^4-x+1)
would it be
(3x^3 - 1)*(f(x^4 - x+1))
 
  • #5
You don't need the equation for [tex]f(x)[/tex].

Ok, let's do the first one first. Start off with the product rule:
[tex](fg)'=f'g+f'[/tex] which can be written more formally as:
[tex](f(x)g(x))'=f'(x)g(x) + f(x)g'(x)[/tex]

The values of [tex]f'(1)[/tex] and [tex]f(1)[/tex] are given, and the values for [tex]g'(1)[/tex] and [tex]g(1)[/tex] can be calculated by plugging [tex]1[/tex] into the [tex]g(x)[/tex] function. And then differentiate [tex]g(x)[/tex] to find [tex]g'(1)[/tex]. Plug in the resulting values into the above equation and you will obtain your solution for the first part.

Finding the derivitive of [tex]g(x)[/tex] should be quite simple.

To do the second part, similarly just plug in your knowns.
 
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  • #6
gotcha thanks
 

1. What is a derivative?

A derivative is a mathematical concept that represents the rate of change of a function at a specific point. It essentially shows how the output of a function changes with respect to its input.

2. How do you calculate derivatives?

The most common method for calculating derivatives is using the rules of differentiation, such as the power rule, product rule, quotient rule, and chain rule. These rules involve manipulating the original function to find the derivative function.

3. Why are derivatives important?

Derivatives have many applications in mathematics, science, and engineering. They are used to find maximum and minimum values of functions, calculate rates of change, and model real-world phenomena. They are also essential in understanding the behavior of complex systems.

4. Can you give an example of calculating derivatives?

Sure, let's say we have the function f(x) = 2x^2. Using the power rule, we can find the derivative as f'(x) = 4x. This means that at any point on the original function, the slope (or rate of change) will be 4 times the value of x.

5. Are there any shortcuts for calculating derivatives?

Yes, there are some special rules and shortcuts for finding derivatives of common functions, such as trigonometric functions, logarithmic functions, and exponential functions. These can save time and make the process easier, but it is important to understand the basic rules of differentiation to use them effectively.

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