Question

what is the value of seceond darivative of sinx(logx)

Answer 1

y= sin(x) log(x)
y'= sin(x)* 1/xln(10) + cos(x)log(x) = (1/ln(10))(sin(x)/x) + cos(x)log(x)
y''= [1/ln(10)][cos(x)*x - (1)sin(x)]/x^2 + [-sin(x)log(x) + cos(x)*[1/xln(10)]
y''= xcos(x)/[x^2ln(10)] - sin(x)/x^2ln(10) - sin(x)log(x) + cos(x)/(xln(10))
y''= cos(x)/(xln(10)) - sin(x)/[x^2ln(10)] - sin(x)log(x) + cos(x)/[xln(10)]
\[y''= 2\cos(x)/[xln(10)] - \sin(x)/[x^2\ln(10)] - \sin(x)\log(x)\]

Answer 2

thanx sir......