How do you find the derivative of #h(p) = (1+p^3) / (5p + (3p^2)#?

Question

Emilia Aguilar · Accepted Answer

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))#

Luke Alvarez · Answer

undefined To find the derivative of $ h(p) = \frac{{1+p^3}}{{5p + 3p^2}} $, use the quotient rule:

$$ h'(p) = \frac{{(5p + 3p^2) \cdot (3p^2) - (1 + p^3) \cdot (5 + 6p)}}{{(5p + 3p^2)^2}} $$