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