Consider the function f(x) for which f(0)=7 and f'(0)=7. Find h'(0) for th function h(x)= 1/f(x). h'(0)= please explain?

since f'(x)=7, if we assume that this is a line 7=m then:

y=7x+b, and since (0,7), b=7 then:

y=7x+7 making

h(x)=1/f(x) become:

h(x)=1/(7x+7)

dh/dx=-7/(7x+7)^2 and when x=0

dh/dx=-1/7

Please read :

h(x) = 1/f(x)

Or for brevity:
h = 1/f

dh/df = -1/f²

dh/dx = (dh/df)(df/dx) (chain rule)
. . .. . .= (-1/f²)df/dx

Writing h'(x) for dh/dx, f'(x) for df/dx and f(x) for f gives:
h'(x) = (-1/f(x)²)f'(x)

For x=0
h'(0) = (-1/f(0)²)f'(0)
. . . . = (-1/7²)7
. . . . = -1/7

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