How do you differentiate #g(x) =e^(1-x)sinhx# using the product rule?

Question

1301 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-23T07:48:31

#g'(x)=(e^(1-x))(coshx-sinhx)#

#color(blue)(y=uv=>(dy)/(dx)=v(du)/(dx)+u(dv)/(dx))#

we have

#g(x)=e^(1-x)sinhx#

#g'(x)=d/(dx)(e^(1-x)sinhx)#

#g'(x)=sinhxd/(dx)(e^(1-x))+(e^(1-x))d/(dx)(sinhx)#

#g'(x)=sinhx(-e^(1-x))+(e^(1-x))coshx#

#g'(x)=(e^(1-x))(coshx-sinhx)#