What is the derivative of #x^(3/2)log_2sqrt(x+1)#?

1 Answer
Mar 13, 2018

#(sqrtx)/(2ln2)[x/(x+1)+3*lnsqrt(x+1)]#

Explanation:

#Note:color(red)((1)log_ab=(log_eb)/(log_ea))#
#color(blue)((2)d/(dx)(u*v)=u*(dv)/(dx)+v*(du)/(dx))#
#y=x^(3/2)log_2sqrt(x+1)=x^(3/2)[(log_esqrt(x+1))/log_e2]#
#y=1/log_e2*x^(3/2)*log_esqrt(x+1)#
#y=1/ln2*x^(3/2)*lnsqrt(x+1)#
#(dy)/(dx)=1/ln2[x^(3/2)d/(dx)(lnsqrt(x+1))+lnsqrt(x+1)d/(dx)(x^(3/2))]#
#=1/ln2[x^(3/2)*1/(sqrt(x+1))*1/(2sqrt(x+1))+lnsqrt(x+1)*3/2*x^(1/2)]#
#=1/ln2[(x*sqrtx)/(2(x+1))+3/2*sqrtx*lnsqrt(x+1)]#
#=(sqrtx)/(2ln2)[x/(x+1)+3*lnsqrt(x+1)]#