where u=xx Differentiating both sides with respect to x, we get ⇒−2‌cos‌e‌c2‌y‌
dy
dx
=(1+‌
1
u2
)‌
du
dx
where u=xx⇒log‌u=x‌log‌x ‌⇒‌
1
u
‌
du
dx
=1+log‌x ‌⇒‌
du
dx
=xx(1+log‌x) ‌∴‌ We get ‌−2‌cos‌e‌c2‌y‌
dy
dx
‌=(1+x−2x)xx(1+log‌x) ‌⇒‌
dy
dx
=‌
(xx+x−x)(1+log‌x)
−2(1+cot2y)
.....(i) Now when x=1,x2x−2xx‌cot‌y−1=0 gives 1−2‌cot‌y−1=0 ⇒cot‌y=0 ∴ From equation (i), at x=1 and cot‌y=0, we get y′(1)=‌