Find the derivative of the function s(t) = √1+sin(t) / 1+cos(t) using the chain rule. s'(t) = ?

Question

Find the derivative of the function s(t) = √1+sin(t) / 1+cos(t) using the chain rule. s'(t) = ?

*the square root covers both the numerator and the denominator*

Answer 1

s(t) = √[(1 + sin t)/(1 + cos t)] = [(1 + sin t)/(1 + cos t)]^½

Below I'm going to use x^1/2 to represent √x (they mean the same thing).

Evaluating s'(t) using the power rule followed by the chain rule

s'(t) = ½[(1 + sin t)/(1 + cos t)]^–1/2 × d[(1 + sin t)/(1 + cos t)]/dt

= ½[(1 + cos t)/(1 + sin t)]^1/2 × d[(1 + sin t)/(1 + cos t)]/dt

Evaluating d[(1 + sin t)/(1 + cos t)]/dt using the quotient rule

d[(1 + sin t)/(1 + cos t)]/dt = [(1 + cos t)(cos t) – (1 + sin t)(–sin t)]/(1 + cos t)² =

[(1 + cos t)(cos t) + (1 + sin t)(sin t)]/(1 + cos t)² =

(cos t + cos²t + sin t + sin²t)/(1 + cos t)² = (1+ cos t + sin t)/(1 + cos t)²

s'(t) = ½[(1 + cos t)/(1 + sin t)]^1/2 × (1+ cos t + sin t)/(1 + cos t)² =

[(1 + cos t)^1/2(1+ cos t + sin t)]/[2(1 + sin t)^1/2(1 + cos t)²] =

(1+ cos t + sin t)/[2(1 + sin t)^1/2(1 + cos t)^3/2]

Answer

s'(t) = (1+ cos t + sin t)/[2(1 + sin t)^1/2(1 + cos t)^3/2]

Answer 2

Let u = (1 + sin(t))/(1 + cos(t))

To find u', we must use the quotient rule:

u' = [cos(t)(1 + cos(t)) - (1 + sin(t))(-sin(t))]/(1 + cos(t))²

u' = [cos(t) + cos²(t) + sin(t) + sin²(t)]/(1 + cos(t))²

using the trig identity that cos²(t) + sin²(t) = 1:

u' = [cos(t) + sin(t) + 1]/(1 + cos(t))²

Substituting "u" in the problem makes it simplify to s(t) = √u or:

s(t) = u^1/2

Using the power rule AND chain rule:

s'(t) = (1/2)u^-1/2•u'

s'(t) = u'/(2u^1/2)

Now just back-substitute the expressions for "u" and u' into u'/(2u^1/2).