" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
Verified answer
Jawaby = [ sin t. tan (t² +1)]²
y = [ p. q]²
y' = 2(pq)(p'q + pq')
p = sin t
p' = cos t
q= tan (t²+1)
q' = (2t +1) sec² (t² +1)
y' = 2 (pq) (p'q + pq')
y' = 2 [sin t. tan(t² +1) { cos t. tan (t²+1) + sin t . (2t+1) sec² (t²+1)}]
y' = 2 [ sin t. tan² (t²+1) cos t + (2t+1) sin² t . sec²(t² +1)]
y' = 2( pq ) ( p'q ) + ( pq' )
q = { 2t + 1 ) sec2 ( T2 + 1 )
y' = 2 ( sin.t.tan ( T2 + 1 ) ( cos t.tan ( T2 + 1 ) + sin.t.( 2t + 1 ) sec2 ( T2 + 1 )))