Simon G. If f and g are the functions whose graphs are shown, let u(x) = f(x)g(x) and v(x) = f(x)/g(x). (a) Find u'(1). =_______ (b) Find v'(5) =_______ More
2 Answers By Expert Tutors
Since u(x) is a product of functions and v(x) is a quotient of functions, why not use the product rule to find u'(1) and the quotient rule to find v'(5)?
William W. answered • 06/29/20
Experienced Tutor and Retired Engineer
Because u(x) = f(x)g(x) then (by the product rule, u'(x) = f '(x)g(x) + f(x)g'(x)
Looking at the graph we see that:
f(1) = 2
f '(1) = 2
g(1) = 1
g'(1) = -1
So u'(1) = (2)(1) + (2)(-1) = 2 - 2 = 0
Apply the same logic for v(x). Since v(x) = f(x)/g(x) then v'(x) = [f '(x)g(x) - f(x)g'(x)]/(g(x))2
f(5) = 3
f '(5) = -1/3
g(5) = 2
g'(5) = 1/3
So v'(5) = [(-1/3)(2) - (3)(1/3)]/(2)2 = (-5/3)/4 = -5/12
Doug C.
Looks like g'(5) = 2/3, i.e. not 1/3. So, -8/12= -2/3.
William W.
You’re right Doug. Thanks.Still looking for help? Get the right answer, fast.
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I'm assuming each square is a 1x1?
Right here is a quick example of the Chain Rule. The basic definition is that the derivative of a function with another function "inside" is that it's the derivative of the "outside" function (leaving the inside function as is) multiplied by the derivative of the "inside" function. So since u(x)=f(g(x)), u'(x)=f'(g(x))*g'(x). Before reading on can you find the derivative of w(x)=g(g(x)?
w'(x)=g'(g(x))*g'(x)
The derivatives at one are g'(1)=-3 and f'(1)=2 (derivatives give you the slope at any particular point, so the slope at one). We also can easily find g(1)=3 and f(1)=2. We just plug in the values to what we found above!
u'(1)=f'(g(1))*g'(1)
=f'(3)*2
looking at the graph, we can see f'(3) is 1/4.
=1/4*2
=1/2
Now can you do the second one on your own? Try before reading on!
w'(1)=g'(g(x)*g'(1)
=g'(3)*-3
looking at the graph, we can see g'(3) is 2/3.
=2/3*-3
=-2
I hope this helps!