Calculus of a Single Variable, Chapter 5, 5.5, Section 5.5, Problem 49

Derivative of a function h with respect to t is denoted as h'(t).
The given function: h(t) = log_5(4-t)^2 is in a form of a logarithmic function.
From the derivative for logarithmic functions, we follow:

d/(dx)log_a(u) =((du)/(dx))/(u*ln(a))

By comparison: log_5(4-t)^2 vs.log_a(u) we should let:
a=5 and u = (4-t)^2
For the derivative of u, recall the Chain Rule formula:
d/(dx)(f(g(x)))= f'(g(x))*g'(x)
Using u=(4-t)^2 , we let:
f(t) = t^2
g(t) = 4-t as the inner function
f'(t)= 2t
f'(g(t))= 2*(4-t)
g'(t)= (-1)
Following the Chain Rule formula, we get:
d/(dx) (4-t)^2= 2 *(4-t)*(-1)
d/(dx) (4-t)^2= -2*(4-t)
or
(du)/(dx)=-2*(4-t)
Plug-in the values:
u =(4-t)^2 , a=5 , and (du)/(dx)=-2*(4-t)
in the d/(dx)log_a(u) =((du)/(dx))/(u*ln(a)) , we get:
d/(dx) (log_5(4-t)^2) = ((-2)*(4-t))/((4-t)^2ln(5))
Cancel out common factor (4-t):
d/(dx) (log_5(4-t)^2) = -2/((4-t)ln(5))
or h'(t)= -2/((4-t)ln(5))