If f and g are the functions whose graphs are shown, let u(x) = f(g(x)), v(x) =
ID: 3027752 • Letter: I
Question
If f and g are the functions whose graphs are shown, let u(x) = f(g(x)), v(x) = g(f(x)), and w(x) = g(g(x)). Find each derivative, if it does not exist, explain why. (If an answer does not exist, enter DNE) u'(1) = If does exist. u'(1) does not exist because f'(1) does not exist. u'(1) does not exist because g'(1) does not exist u'(1) does not exist because f'(3) does not exist. u'(1) does not exist because g'(2) does not exist. v'(1) = If does exist. v'(1) does not exist because f'(1) does not exist. v'(1) does not exist because g'(1) does not exist v'(1) does not exist because f'(3) does not exist. v'(1) does not exist because g'(2) does not exist. w'(1) = If does exist. w'(1) does not exist because f'(1) does not exist. w'(1) does not exist because g'(1) does not exist w'(1) does not exist because f'(3) does not exist. w'(1) does not exist because g'(2) does not exist.Explanation / Answer
u(x) = f( g(x) )
=> u'(x) = f'( g(x) ) . g'(x)
v(x) = g( f(x) )
=> v'(x) = g'( f(x) ) . f'(x)
w(x) = g( g(x) )
=> w'(x) = g'( g(x) ) . g'(x)
(a) Calculating the required values
=> u'(1) = f'( g(1) ) . g'(1)
=> u'(1) = f'( g(1) ) . g'(1)
=> u'(1) = f'( 3 ) . ( -3 )
=> u'(1) = ( -1/4 ) . ( -3 ) = 3/4
(b) Calculating the required values
=> v'(1) = g'( f(1) ) . f'(1)
=> v'(1) = g'( 2 ) . f'(1)
Since , there is a kink at x = 2 for g(x) , hence g'(2) does not exist , therefore , v'(1) does not exist
(c) Calculating the required values
=> w'(1) = g'( g(1) ) . g'(1)
=> w'(1) = g'( 3 ) . g'(1)
=> w'(1) = ( 2/3 ) . ( -3 ) = -2
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