How do you find the derivative of # y = (x ln(x)) / sin(x)#?

1 Answer
Feb 13, 2016

#( sinx(1+ lnx) - cosx. xlnx)/sin^2x#

Explanation:

using the#color(blue)(" quotient rule ") #

If f(x) #= g(x)/(h(x))#

then f'(x)# =( h(x)g'(x) - g(x)h'(x) )/(h(x))^2 #

now: xlnx is a product of 2 functions and as such will require the#color(blue)(" product rule ") #

If f(x) = g(x)h(x) then f'(x) = g(x)h'(x) + h(x)g'(x)
#color(black)("-----------------------------------------------")#
differentiating the numerator using 'product rule"

# x d/dx(lnx) + lnx d/dx(x) = x.(1/x) + lnx .1= 1 + lnx #

differentiate the denominator: # d/dx(sinx) = cosx #
#color(black)("---------------------------------------------")#

back to the original function using the 'quotient rule'

#dy/dx = (sinx d/dx(xlnx) - xlnx d/dx(sinx))/sin^2x #

'inserting' the derivates gives

# dy/dx =( sinx(1 + lnx) - cosx.xlnx)/sin^2x #