Find dy/dx of #y=e^(x sin x) # Help please and show the working? Thanks

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Answer 1 · 2017-12-22T14:05:22

#dy/dx=e^(x*sin(x))*(sin(x)+x*cos(x))#

#y=e^(x*sin(x))#

Use the chain rule #dy/dx=(du)/(dx)*(dy)/(du)#

Let #u=x*sin(x)# and #y=e^u#

Then #(du)/dx=sin(x)+x*cos(x)# and #dy/(du)=e^u#

#dy/dx=(sin(x)+x*cos(x))*e^u#

Substitute #u=x*sin(x)#

#dy/dx=e^(x*sin(x))*(sin(x)+x*cos(x))#

Answer 2 · 2017-12-22T14:11:53

The answer is #=e^(xsinx)(sinx+xcosx)#

Reminder (the chain rule and the product rule)

#(e^(u(x)))'=e^(u(x))* u'(x)#

#(u(x)v(x))'=u'(x)v(x)+u(x)v'(x)#

Here

#u(x)=xsinx#

#u'(x)=(x)'sinx+x(sinx)'=sinx+xcosx#

Therefore,

#dy/dx=e^(xsinx)(sinx+xcosx)#