course mth 271
mr smith: im still not getting this material. also, please dont forget to send me a run down of the quiz. for some reason the email i sent you came back mailer demon. thanks, shelley
f ' (g(x)) is .5 e^(x)^-.5 = .5 e^(-.5 x)
(f(g(x))) ' = g ' (x) * f ' (g(x)) = e^x * ( .5 e^(-.5 x)).
(f(g(x))) ' includes e^x, which comes from g ' (x). This term is not part of f ' (g(x)).
i still dont follow. does f'(g)x) mean the deriv of f(x) times g(x). or do you plug g(x) in to the deriv of f(x)?
###f ' ( z) would take the value of f ' at z. We know that f ' (z) = .5 e^(-.5 x).
What would be each of the following:
f ' (a - b)
f ' (x + h)
f ' (aardvark)
f ' (cos(x))
f ' (p(x))
f ' (g(x))?
The chain rule says that you first identify the f and g functions, with g being the 'inner' function, the first function encountered by the variable. f is the function into which this result will be substituted. The function f does not act on the variable (as it would in a product or quotient rule) but on the g(x) function.
For this reason the f ' function doesn't act on the variable, but on the g(x) function. So f ' (g(x)) means exactly what it says; substitute g(x) for the variable in the f ' function. However f ' (g(x)) is not the whole picture. This function does not give you the derivative of (f(g(x))) because the expression f ' (g(x)) does not account for the rate-of-change behavior of g(x).
Since the f function acts on the g(x) function, the derivative of the composite function f(g(x)) must account for the fact that the variable x has some rate of change first due to the g function, then when f comes along (acting on the g) function the f function contributes also the rate of change of the composite function.
We account for the rate-of-change behavior of g(x) by mulitplying g ' (x) by f ' (g(x)). g ' (x) is the rate of change of g with respect to x, evaulted at our original variable x. f ' (g(x)) is the derivative of the f function, evaluated not as x but at g(x). If we want the rate of change of the composite function f(g(x)) we have to multiply the contribution g ' (x) by the contribution f ' (g(x)).
For this reason we express the chain rule as (f(g(x))) ' = g ' (x) * f ' (g(x)).
Thus (f(g(x))) ' = g ' (x) * f ' (g(x)) = e^x * ( .5 e^(-.5 x)).
The e^x term is the g ' (x).
The .5 e^(-.5 x)) terms is the f ' (g(x)).###
Here g(x) = sqrt(x) and f(z) = e^z so g ' (x) = 1 / (2 sqrt(x)) or .5 x^-.5, and f ' (z) = e^z. So
(f(g(x))) ' = g ' (x) * f ' (g(x)) = .5 x^-.5 * e^(sqrt(x)).&& i really dont follow here
If f ' (z) = e^z, then what is f ' (g(x))?
g ' (x) = 1 / (2 sqrt(x)) so what is (f(g(x))) ' = g ' (x) * f ' (g(x))?
You have f ' (z) = e^z, g(x) = sqrt(x), g ' (x) = 1 / (2 sqrt(x)).
You have everything you need to plug into the formula
(f(g(x))) ' = g ' (x) * f ' (g(x)).
What do you get?###
I dont have any idea where to start w/ these problems..
e^(1/x)
ln ( 1 / x^2 )
sqrt(ln(x)).
###To start each problem, verify that each is in fact a composite.
Then identify the two functions f(z) and g(x).
The find the derivatives of f(z) and g(x).
Then plug into the chain rule statement (f(g(x))) ' = g ' (x) * f ' (g(x)).###
See if my additional notes help. If not, let me know and send more questions.