How do you differentiate #y=(sinx)^lnx#?
1 Answer
Feb 21, 2017
# dy/dx = (sinx)^(lnx){(lnx)(cotx) + (ln(sinx))/x} #
Explanation:
Generally when dealing with a variable exponent it is easier to differentiate (and understand) by taking natural logarithms (to remove the exponent) and differentiating implicitly:
We have:
# y = (sinx)^(lnx) #
Take Natural logarithms:
# lny = ln{(sinx)^(lnx)} #
# \ \ \ \ \ \= (lnx)(ln(sinx)) \ \ \ # (rule of logs)
Differentiate wrt
# \ 1/ydy/dx = (lnx)(1/sinx*cosx) + (1/x)(ln(sinx)) #
# \ \ \ \ \ \ \ \ \ \ \= (lnx)(cotx) + (ln(sinx))/x #
# :. dy/dx = y{(lnx)(cotx) + (ln(sinx))/x} #
# \ \ \ \ \ \ \ \ \ \ = (sinx)^(lnx){(lnx)(cotx) + (ln(sinx))/x} #
