If y- T ax T anx Tan x then dy - dx at x -4 isA- 3B- 4C- 1D- 2

The correct option is D 2log y=(tan x)^tan xlog (tan x)→(1) Taking log again, we get from (1) log(log y)=tan x log (tan x)+log(log(tan x)) Differentiate ...

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Byju's Answer

Standard XII

Mathematics

Integration by Partial Fractions

If y= T ax T ...

Question

$I f y = (T a x)^{(T a n x)^{T a n x}} t h e n \frac{d y}{d x} a t x = \frac{π}{4} i s$

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Solution

The correct option is B 2
$l o g y = (t a n x)^{t a n x} l o g (t a n x) \to (1) T a k i n g l o g a g a i n, w e g e t f r o m (1) l o g (l o g y) = t a n x l o g (t a n x) + l o g (l o g (t a n x)) D i f f e r e n t i a t e w i t h r e s p e c t t o x \frac{1}{l o g y} . \frac{1}{y} \frac{d y}{d x} = {sec}^{2} x l o g (T a n x) \frac{1}{T a n x} S e c^{2} x \frac{1}{l o g (T a n x)} \frac{1}{T a n x} S e c^{2} x$ $∴ \frac{d y}{d x} = y l o g y s e c^{2} x (l o g (T a n x) + 1 + \frac{1}{T a n x l o g (T a n x)})$ $= y (T a n x)^{T a n x} l o g T a n . S e c^{2} x [(l o g (T a n x) + 1) + \frac{1}{T a n x l o g (T a n x)}] = y (T a n x)^{T a n x} S e c^{2} x [l o g (T a n x) (l o g T a n x + 1) + C o t x] W h e n a t x = \frac{π}{4}, y = 1 ∴ \frac{d y}{d x} = 1.1.2 (0 + 1) = 2$

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If y=(Tax)(Tanx)Tanxthen dydx at x=π4 is

$I f y = (T a x)^{(T a n x)^{T a n x}} t h e n \frac{d y}{d x} a t x = \frac{π}{4} i s$