course mth 271
First I want to see if you understand function notation: If f(z) = z^2, then what are the values of f(2), f(7), f(a + h), f(q - p), f(aardvark) and f(g(x))?
f(2)=4, f(7)=49 f(a+h)= (a+h)^2 f(q-p)=(q-p)^2 f(aard)=(aard)^2
f(g(x))=(g(x))^2
Next I want you to think through, step by step, what you would do to evaluate a series of expressions:
If you are given the value of x, then if you want to find the value of sqrt(x) * e^x, what is the first thing you would do? What is the next thing you would do? What further steps, if any are necessary, would you take to get the value of the expression?
plug in x. eval sqrootx, eval e^x, multiply answers
If you are given the value of x, then if you want to find the value of (x - 3)^2, what is the first thing you do with that value of x? What is the next thing? What further steps, if any are necessary, would you take to get the value of the expression?
plug in x, subtract, square
If you are given the value of x, then if you want to find the value of (e^x) / (sqrt(x)), what is the first thing you do with that value of x? What is the next thing? What further steps, if any are necessary, would you take to get the value of the expression?
plug in x, eval e^x, eval sqrootx, divide
In the above, in one case you evaluated two different functions for x then multiplied the results (#2), in another you evaluated two different functions for x and divided the result (#4), and in another you evaluated one function for x and then plugged the resulting value into another function to get your final result(#3). Which was which? In each case there were two functions involved. What were the two functions for each?
2) f(x)=sq rootx g(x)= e^x
3) f(x)=(x-3) g(x)= (f(x))^2
**If you let g(z) = z^2, then it follows that g(x) = (f(x))^2 (very similar to the first series of questions).
So you could say that (x-3)^2 = g(f(x)), where f(x) = x-3 and g(x) = x^2.
So the 'inner' function is the x - 3 function; this is the first function you evaluate when you are given a value of x. If we call the resulting value z, is then 'fed into' the 'outer' function g(z) = f(z^2) to get the final value. So again, the function is g(f(x)), where f(x) is the 'inner' function and g(z) the 'outer'.
Your thinking is correct, but you are going to want to reverse your usage of f and g so that g is the 'inner' function and f the 'outer' function. This is because the chain rule is usually stated in terms of f(g(x)), not g(f(x)), and unless you use a notation consistent with the usual usage you will get confused.
So instead of letting g(z) = z^2 and f(x) = x - 3, just reverse the names so that f(z) = z^2 and g(x) = x - 3. Now the function is in the form f(g(x)).
The derivatives of these functions are f'(x) = 2 x, using the power rule, and g ' (x) = (x) ' - (3) ' = 1 - 0 = 1.
So what is f ' (g(x)) ?
What therefore is the derivative (f(g(x)) ' = g ' (x) * f ' (g(x))?**
4)f(x)= e^x g(x)=sqrootx
Each of the following involves either evaluating two different functions for x then multiplying the results, evaluating two different functions for x and dividing the result, or evaluating one function for x and then plugging the resulting value into another function to get your final result. For each, identify the two functions involved and state what you did with those two functions:
sqrt(x) * ln(x)
f(x)=sqrootx g(x)= lnx
plug in x, eval sqrtx, eval ln(x), multiply
**
OK. You have correctly identified this as a product of two functions and you have identified the functions f and g.
For your functions, what is f ' , what is g ' and what therefore is
(f g) ' = f ' g + g ' f?
**
e^(x^2)
f(x)=x^2 g(x)=e^(f(x))
plug in x, eval x^2, eval e^answer
**
If g(x) = e^(f(x)), then g(f(x)) would be found by replacing x with f(x), so g(f(x)) would be e^(f(f(x))). This is not what you intend.
In any case, reverse your notation so that g is the 'inner' function. You will have
g(x)=x^2, f(z)=e^z.
We don't want to write f(x) = e^(g(x)) because we want to get the function into the form f(g(x)). If f(x) = e^(g(x)) then f(g(x)) would replace x by g(x) and we would have f(g(x)) = e^(g(g(x))). We want f(g(x)) = e^(g(x)), not e^(g(g(x))).
If we write f(z) = e^z, then f(g(x)) means e^(g(x)), as we intend.
So what is f ' (z) and what is g ' (x)?
What therefore is (f(g(x))) ' = g ' (x) * f ' (g(x))?
**
(x^2-5) / ln(x)
plug in x, eval num, eval denom, divide
f(x)=(x^2-5) g(x)=ln(x)
**
Good. You have identified that this is a quotient of two functions f and g.
What are f ' and g '?
What therefore is (f / g) / = (f ' g - g ' f) / g^2?
**
sqrt(e^x)
f(x)=e^x g(x)=sqroot(f(x))
plug in x, eval e^x, take sqroot
**
Right idea but use g for the 'inner' function, and use the 'dummy' variable z to write the 'outer' function:
g(x) = e^x, f(z) = sqrt(z).
Find g ' and f ' then apply the chain rules.
**
(x^3 - 4 x + 1) * e^x
plug in x, eval e^x, distribute answer, combine like terms
f(x)=(x^3-4x+1) g(x)=e^x
**
Good. Is this a product, a quotient or a composite? What rule applies? What therefore is the derivative? (If it's a composite fix your notation so that g(x) is the 'inner' function and f is expressed in terms of a 'dummy' variable (e.g., z, and I have used it, or u, as I believe you text uses it; it doesn't matter what letter you use for your 'dummy' variable as long as it isn't used elsewhere in the problem for a different variable; then apply the chain rule).
**
e^(sqrt(x)
plug in x, take sqroot, eval e^answer
f(x)=sqroot(x) g(x)=e^(f(x))
**
Good. Is this a product, a quotient or a composite? What rule applies? What therefore is the derivative? (If it's a composite fix your notation so that g(x) is the 'inner' function and f is expressed in terms of a 'dummy' variable, etc.).
**
x^2 / e^x
plug in x, eval num, eval denom, divide
f(x)=x^2, g(x)=e^x
**
Good. Is this a product, a quotient or a composite? What rule applies? What therefore is the derivative? (If it's a composite fix your notation so that g(x) is the 'inner' function and f is expressed in terms of a 'dummy' variable, etc.).
**
You're seeing the functions correctly. Your notation on the chain rule needs to be reversed so that g is the 'inner' function, and you have to use a 'dummy' variable to express the 'outer' function f. See my notes for details, and answer the subsequent questions I've added.