`h(t) = log_5(4-t)^2` Find the derivative of the function

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))`