M Marcia New member Joined Oct 18, 2005 Messages 14 Oct 18, 2005 #1 Suppose that f, g are differentiable functions and g(1)=3 , g'(1)=-4 , f'(3)=5. Let h(x)=f(g(x)). Find h'(1) h'(x) = f(x)g'(x)+g(x)f'(x) Is this even a start?
Suppose that f, g are differentiable functions and g(1)=3 , g'(1)=-4 , f'(3)=5. Let h(x)=f(g(x)). Find h'(1) h'(x) = f(x)g'(x)+g(x)f'(x) Is this even a start?
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Oct 18, 2005 #2 Hello, Marcia! Suppose that f, g are differentiable functions and g(1) = 3 , g'(1) = -4 , f '(3) = 5. Let h(x) = f(g(x)). Find h'(1) h'(x) = f(x)g'(x) + g(x)f'(x) . . . Is this even a start? . . . . . sorry, no . . . it's not a product. Click to expand... The function is a composite: .h(x) .= .f(g(x)) . . . f of g(x) We use the Chain Rule: . h'(x) .= .f '(g(x))·g'(x) Now we want: .h'(1) .= .f '(g(1))·g'(1) . . . . we're told: .g(1) = 3, g'(1) = -4 . . . . . . . . . . . . . . . . . . . . . . .↓. . . . ↓ . . . . . . . . . . . . .h'(1) .= .. f '(3) · (-4) . . . . we're told: .f '(3) = 5 . . . . . . . . . . . . . . . . . . . . . . .↓ . . . . . . . . . . . . .h'(1) .= . . . 5 · (-4) . = . -20
Hello, Marcia! Suppose that f, g are differentiable functions and g(1) = 3 , g'(1) = -4 , f '(3) = 5. Let h(x) = f(g(x)). Find h'(1) h'(x) = f(x)g'(x) + g(x)f'(x) . . . Is this even a start? . . . . . sorry, no . . . it's not a product. Click to expand... The function is a composite: .h(x) .= .f(g(x)) . . . f of g(x) We use the Chain Rule: . h'(x) .= .f '(g(x))·g'(x) Now we want: .h'(1) .= .f '(g(1))·g'(1) . . . . we're told: .g(1) = 3, g'(1) = -4 . . . . . . . . . . . . . . . . . . . . . . .↓. . . . ↓ . . . . . . . . . . . . .h'(1) .= .. f '(3) · (-4) . . . . we're told: .f '(3) = 5 . . . . . . . . . . . . . . . . . . . . . . .↓ . . . . . . . . . . . . .h'(1) .= . . . 5 · (-4) . = . -20
