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 :)
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?
it is assumed that f is differentiable and that w has an absolute maximum at t0
Show that f(t0) = f′(t0)(C + t0).
To find the derivative of w(t), you can use the quotient rule. The quotient rule states that if you have a function u(t) divided by v(t), then the derivative is given by (u'(t)v(t) - u(t)v'(t)) / [v(t)]^2.
In this case, u(t) = f(t) and v(t) = c + t. Taking the derivatives, we have:
u'(t) = f'(t) (the derivative of f(t) with respect to t)
v'(t) = 1 (the derivative of (c + t) with respect to t is just 1)
Applying the quotient rule, we get:
w'(t) = [f'(t)(c + t) - f(t)(1)] / [(c + t)]^2
= [f'(t)(c + t) - f(t)] / [(c + t)]^2
To prove that f(t_0) = f'(t_0) (C+t_0), you need to set t equal to t_0 in the expression for w'(t) and simplify:
w'(t_0) = [f'(t_0)(c + t_0) - f(t_0)] / [(c + t_0)]^2
Now, to simplify this expression, we can multiply the numerator and denominator by (c + t_0)^2:
w'(t_0) = [f'(t_0)(c + t_0) - f(t_0)] / [(c + t_0)(c + t_0)]
= [f'(t_0)(c + t_0) - f(t_0)] / [c^2 + 2c(t_0) + (t_0)^2]
Now, if f(t_0) = f'(t_0)(C+t_0), we can substitute this equality into the expression for w'(t_0):
w'(t_0) = [f'(t_0)(c + t_0) - f(t_0)] / [c^2 + 2c(t_0) + (t_0)^2]
= [f'(t_0)(c + t_0) - f'(t_0)(C+t_0)] / [c^2 + 2c(t_0) + (t_0)^2]
= 0
Therefore, if f(t_0) = f'(t_0)(C+t_0), then the derivative w'(t_0) will be zero.