Question -17

Calculus Level 3

Let $y$ be an implicit function of $x$ defined by $x^{2x}-2 x^x \cot (y)-1=0$ then find the value of $y'(1)$ .

Note : $y'(x)=\dfrac{\mathrm{d}y}{\mathrm{d}x}$

Looking for Top Rank in JEE ? Alright, then go for this set : Expected JEE-Mains-2015-B .

-1

-log2

log2

1 solution

Rahul Singh
Mar 4, 2015

The equation can be written as.. y=cot^(-1)((x^x-1).(x^x+1)/2x^x) Now in order to simplify it,let x^x=t such that when x=1,we have t=1. the simplified equation will be.. y=cot^(-1)((t-1).(t+1)/2t) Differentiate this w.r.t 't' & put t=1 to get the solution '-1' .

(x^x)^2 - 2x^x cot(y) - 1 = 0

x^x = (2 cot(y) + sqrt{4 cot^2(y) + 4})/2

x^x = (2 cot(y) + 2 sqrt{cot^2(y) + 1})/2

x^x = cot(y) + csc(y)

i get stuck!

Muhammad Ihsan - 6 years, 1 month ago

Question -17

Looking for Top Rank in JEE ? Alright, then go for this set : Expected JEE-Mains-2015-B .

1 solution

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