Asked by Mable
w(t)= f(t)/(c+t)
w'(t)=?
I got f'(t)(c+t)-f(t)/(c+t)^2 as the derivative and I'm having a hard time trying to prove that f(t_0) = f'(t_0) (C+t_0)
Help is always appreciated :)
w'(t)=?
I got f'(t)(c+t)-f(t)/(c+t)^2 as the derivative and I'm having a hard time trying to prove that f(t_0) = f'(t_0) (C+t_0)
Help is always appreciated :)
Answers
Answered by
Bosnian
Correct answer?
Quotient Rule for Derivatives:
[ f ( t ) / g ( t ) ] ´ = g ( t ) * f ´( t ) - f ( t ) * g ´( t ) / g ( t ) ^ 2
In this case g ( t ) = c + t so:
w ( t )´ = [ ( c + t ) * f´( t ) - f ( t ) * ( c + t )´ ] / ( c + t ) ^ 2 =
[ ( c + t ) * f´( t ) - f ( t ) * ( 0 +1 ) ] / ( c + t ) ^ 2 =
[ ( c + t ) * f´( t ) - f ( t ) * 1 ] / ( c + t ) ^ 2 =
[ ( c + t ) * f´( t ) - f ( t ) ] / ( c + t ) ^ 2
What (t_0) = f'(t_0) (C+t_0) mean?
Quotient Rule for Derivatives:
[ f ( t ) / g ( t ) ] ´ = g ( t ) * f ´( t ) - f ( t ) * g ´( t ) / g ( t ) ^ 2
In this case g ( t ) = c + t so:
w ( t )´ = [ ( c + t ) * f´( t ) - f ( t ) * ( c + t )´ ] / ( c + t ) ^ 2 =
[ ( c + t ) * f´( t ) - f ( t ) * ( 0 +1 ) ] / ( c + t ) ^ 2 =
[ ( c + t ) * f´( t ) - f ( t ) * 1 ] / ( c + t ) ^ 2 =
[ ( c + t ) * f´( t ) - f ( t ) ] / ( c + t ) ^ 2
What (t_0) = f'(t_0) (C+t_0) mean?
Answered by
Mable
it is assumed that f is differentiable and that w has an absolute maximum at t0
Show that f(t0) = f′(t0)(C + t0).
Show that f(t0) = f′(t0)(C + t0).
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