How do you differentiate #F(y)=(1/y^2-3/y^4)(y+5y^3)#?

Question

6528 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-19T14:24:39

#d/dyF(y)=15y^2+59/y^2+9/y^4-55#

#F(y)=(1/y^2-3/y^4)(y+5y^3)#

Since, #d/dxg(x)f(x)=g(x)f'(x)+f(x)g'(x)#

#d/dyF(y)=(1/y^2-3/y^4)d/dy(y+5y^3)+d/dy(1/y^2-3/y^4)(y+5y^3)#

#d/dyF(y)=(1/y^2-3/y^4)(1+15y^4)+(-2/y^3+12/y^5)(y+5y^3)#

#d/dyF(y)=(1/y^2-3/y^4-45+15y^2)+(-2/y^2+12/y^4-10+60/y^2)#

#d/dyF(y)=1/y^2-3/y^4-45+15y^2-2/y^2+12/y^4-10+60/y^2#

#d/dyF(y)=15y^2+59/y^2+9/y^4-55#