Find dy/dx, when y = (tan x)^(cot x) + (cot x)^(tan x)

Question

Find \(\cfrac{dy}{d\text x}\), when

y = (tan x)^{(cot x)} + (cot x)^{(tan x)}

Answer 1

Let y = (tan x)^{(cot x)} + (cot x)^{(tan x)}

⇒ y = a + b

where a= (tan x)^{cot x} ; b = (cot x)^{tan x}

  \(\Bigg\{\) Using chain rule, \(\cfrac{d(u +a)}{d\text x}=\cfrac{du}{d\text x}+\cfrac{da}{d\text x}\) where a and u are any variables \(\Bigg\}\)

a= (tan x)^{cot x}

Taking log both the sides:

⇒ log a= log (tan x) cot x

⇒ log a= cot x log (tan x)

{log x^a = a log x}

Differentiating with respect to x:

Put the value of a = (tan x)^{cot x} :

⇒ \(\cfrac{da}{d\text x}\) = (tan x)^{cot x}{cot² x sec²x - cosec²x log(tan x)}

b = (cot x)^{tan x}

Taking log both the sides:

⇒ log b= log (cot x)^{tan x}

⇒ log b= tan x log (cot x)

{log x^a = a log x}

Differentiating with respect to x: