What will be the differentiation ofthe [log tan t/2] ????

We need to differentiate the function:

f(t) = log(tan(t/2))

Step 1: Apply the Chain Rule
Let u = tan(t/2), so that we rewrite the function as:

f(t) = log(u)

Now, differentiate both sides:

df/dt = (1/u) * du/dt

Step 2: Differentiate u = tan(t/2)
We use the derivative of tan(x):

d/dt [tan(t/2)] = sec²(t/2) * (1/2)
= (1/2) sec²(t/2)

Step 3: Substitute back into the expression
Now, we substitute du/dt into df/dt:

df/dt = (1/tan(t/2)) * (1/2) sec²(t/2)

Using the identity sec²(x) = 1 + tan²(x), we rewrite:

df/dt = (1/2) * (sec²(t/2) / tan(t/2))
= (1/2) * ((1 + tan²(t/2)) / tan(t/2))
= (1/2) * (cot(t/2) + tan(t/2))

Final Answer:
df/dt = (1/2) * [cot(t/2) + tan(t/2)]