IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

vinu

Hiya ppl...

y = Tan^-1 [ {log(e/2)/x} / log(ex²) ] ;

Find d²y/dx² .

KAB

In the numerator is it {log(e/2)}/x ???

taruntanuj007

y = arctan[ {(1 - log2)/x } / (1 + 2logx) ]

this is the most sim[plified version of the expression u gave so i am differentiating

if this is right u can easily differentiate it might be a bit long but no problem differentiating it..........if ur q is right and my xplananion then nudge me!!!!

gauravgk

y = Tan^-1 [ {log(e/2)/x} / log(ex²) ] ;

Find d²y/dx² .

y=tan^-1[{log(e/2)-log(x)}-{log(e)+2logx}]

y=tan^-1[{log(e)-log(2)-log(x)}-{log(e)+2logx}]

y=tan^-1[1-log2-logx-1-2logx]

y=tan^-1[log2-logx-2logx]

y=tan^-1[log2-3logx]

y=tan^-1[log{2/x³]

solve it.....

Ask & Discuss Questions with Community & Experts