How do you differentiate #g(x) = sqrt(e^x-x)cosx# using the product rule?

Question

1486 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-01T02:09:23

Say #y=g(x)#

This means that:

#y=sqrt(e^x-x)*cosx#

#y=u*v#

Whereby both u and v are functions of x .

If this is the case, then:

#(dy)/(dx)=u*(dv)/(dx)+v*(du)/(dx)=g'(x)#

#u=sqrt(e^x-x)#

#u^2=e^x-x#

#2u*(du)/(dx)=e^x-1#

#(du)/(dx)=(e^x-1)/(2u)#

#(du)/(dx)=(e^x-1)/(2sqrt(e^x-x))#

Using normal differentiation...

#v=cosx#

#(dv)/(dx)=-sinx#

This means that:

#(dy)/(dx)=-sinxsqrt(e^x-x)+cosx*(e^x-1)/(2sqrt(e^x-x))#

Therefore:

#g'(x)=-sinxsqrt(e^x-x)+cosx*(e^x-1)/(2sqrt(e^x-x))#