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Question Number 81565 by jagoll last updated on 14/Feb/20

I f f (x) = {(\tan x)}^{\cot x} + {(\cot x)}^{\tan x}

f^{'} (x) = ?

Commented by john santu last updated on 14/Feb/20

good question

Commented by mathmax by abdo last updated on 14/Feb/20

f (x) = {(t a n x)}^{c o t a n x} + {(c o t a n x)}^{t a n x} \Rightarrow

f (x) = e^{\frac{1}{t a n x} l n (t a n x)} + e^{t a n x l n (\frac{1}{t a n x})} = e^{\frac{l n (t a n x)}{t a n x}} + e^{- t a n x l n (t a n x)}

= u (x) + v (x) \Rightarrow f^{^{'}} (x) = u^{^{'}} (x) + v^{^{'}} (x)

u^{^{'}} (x) = {(\frac{l n (t a n x)}{t a n x})}^{^{'}} u (x) = {\frac{(1 + {t a n}^{2} x)}{t a n x} \times t a n x - l n (t a n x) (1 + {t a n}^{2} x)} \times \frac{1}{{t a n}^{2} x} \times u (x)

= {1 + {t a n}^{2} x - l n (t a n x) - l n (t a n x) {t a n}^{2} x} \times \frac{u (x)}{{t a n}^{2} x}

v^{^{'}} (x) = {(- t a n x l n (t a n x))}^{^{'}} v (x)

= - {(1 + {t a n}^{2} x) l n (t a n x) + t a n x \times \frac{1 + {t a n}^{2} x}{t a n x}} v (x)

= - {(1 + {t a n}^{2} x) l n (t a n x) + 1 + {t a n}^{2} x} v (x)

Answered by john santu last updated on 14/Feb/20

Commented by jagoll last updated on 14/Feb/20

t h a n k a l o t o f

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