BITSAT Mathematics Question (2023) — Solution

Step-by-step solution

Here, y ^ x = e ^ y - x Taking log on both sides, we get aligned & y^x= e^ y-x \\ & ( a^b=b a and e=1 ) aligned x y=(y-x) e x y=y-x (i) On differentiating w.r.t. x, we get d d x (x y)= d d x (y-x) (using product rule) aligned & x ( 1 y ) d y d x + y(1)= d y d x -1 \\ & d y d x ( x y -1 )=-1- y \\ & d y d x [ y (1+ y) y -1 ]=-(1+ y) \\ & [ from eq. (i) x= y (1+ y) ] \\ & d y d x [ 1-1- y 1+ y ]=-(1+ y) aligned d y d x =- (1+ y)^2 - y d y d x = (1+ y)^2 y