Here, we can use the product rule, which states that, be #y=f(x)g(x)#, then:
#y'=f'(x)g(x)+f(x)g'(x)#
So, rewriting our quotient, we have #(1+p^3)(5p+3p^2)^-1#.
To derivate the denominator, we'll use the chain rule, which states that #(dy)/(dx)=(dy)/(du)(du)/(dx)#. Here, we'll rename #u=5p+3p^2#.
Let's go!
#(dh(p))/(dp)=(3p^2)(5p+3p^2)^-1+(1+p^3)(-1)(5+6p)(5p+3p^2)^-2#
#(dh(p))/(dp)=(3p^2)/(5p+3p^2)-(5+5p^3+6p+6p^4)/(5p+3p^2)^2#
l.c.d: #(5p+3p^2)^2#
#(dh(p))/(dp)=((3p^2)(5p+3p^2)-(5+5p^3+6p+6p^4))/(5p+3p^2)^2#
#(dh(p))/(dp)=(15p^3+9p^4-5-5p^3-6p-6p^4)/(5p+3p^2)^2#
#(dh(p))/(dp)=(3p^4+10p^3-6p-5)/(25p^2+30p^3+9p^4)#
#(dh(p))/(dp)=(3p^4+10p^3-6p-5)/(p^2(25+30p+9p^2))#
We can factor the denominator's parenthesis by finding its roots:
#(-30+-sqrt(900-4(9)(25)))/18#
#p_1=p_2=-5/3#, thus #(3p+5)(3p+5)=(25+30p+9p^2)#
Final answer:
#(dh(p))/(dp)=color(green)((3p^4+10p^3-6p-5)/(p^2(3p+5)^2))#
