Question #dfcde

1 Answer
Nov 4, 2017

See the following :)

Explanation:

As this question contain some complicated degree, we can use logarithmic differentiation to slove this problem.

First, we take #ln# on both sides of the equation.

#y=(sinx)^lnx*(lnx)^cosx#

#ln(y)=ln[(sinx)^lnx*(lnx)^cosx]#

#lny=ln[(sinx)^lnx]+ln[(lnx)^cosx]#

#lny=lnx*ln(sinx)+ cosx*ln(lnx)#

Now, the problem of complicated degree are solved and we can differentiate both sides.

#d/dx(lny)=d/dx[lnx*ln(sinx)+ cosx*ln(lnx)]#

#1/y*dy/dx=d/dx[lnx*ln(sinx)]+ d/dx[cosx*ln(lnx)]#

#1/y*dy/dx=(lnx)*d/dx[ln(sinx)]+ln(sinx)*d/dx(lnx)+ cosx*d/dx[ln(lnx)]+ln(lnx)*d/dx(cosx)#

#1/y*dy/dx=(lnx)*1/sinx*cosx+ln(sinx)*(1/x)+ cosx*1/(lnx)*d/dx(lnx)+ln(lnx)*(-sinx)#

#1/y*dy/dx=(lnxcosx)/(sinx)+(ln(sinx))/x+ cosx*1/(lnx)*1/x-sinx ln(lnx)#

#1/y*dy/dx=(lnxcosx)/(sinx)+(ln(sinx))/x+ cosx/(xlnx)-sinx ln(lnx)#

#dy/dx=[(lnxcosx)/(sinx)+(ln(sinx))/x+ cosx/(xlnx)-sinx ln(lnx)]*y#

#dy/dx=[(lnxcosx)/(sinx)+(ln(sinx))/x+ cosx/(xlnx)-sinx ln(lnx)]*[(sinx)^lnx*(lnx)^cosx]#

Here is the answer. Hope this can help you :)
If you need further explaination, feel free to ask me.