can anyone help me find the derivative of p(x) if p(x) = ( x3+2 x+4 ) ( 2 x2+x+5 )?

Answer 1

p(x) = (x³ + 2x + 4)(2x² + x + 5)
p `(x)
= (3x² + 2)(2x² + x + 5) + (4x + 1) (x³ + 2x + 4)
= 6x^(4) + 3x³ + 19x² + 2x + 10
+ 4x^(4) + x³ + 8x² + 18x + 4

=10x^4 + 4x³ + 27x² + 20x + 14

Answer 2

Rule 5) To find the derivative of the product of two functions, take the first times the derivative of the second and add it to the second times the derivative of the first.
p(x) = (x^3 + 2x + 4)(2x^2 + x + 5)
Derivative of p(x) = (x^3 + 2x + 4)(4x + 1) + (2x^2 + x + 5)(3x^2 + 2)
expand
4x^4 + 8x^2 +16x + x^3 + 2x + 4 + 6x^4 + 3x^3 + 15x^2 + 4x^2 + 2x + 10
then combine/simplify terms
4x^4 + 6x^4 + x^3 + 3x^3 + 8x^2 + 15x^2 + 4x^2 + 16x + 2x + 2x + 4 + 10
10x^4 + 4x^3 + 27x^2 + 20x + 14

Answer 3

This can be solved by two ways

First one

p(x)=(x^3+2x+4)(2x^2+x+5)
If you open the bracket and multiply the terms the equation is in the form of

p(x)=2x^5+x^4+9x^3+10x^2+14x+20
Its derivative is
p'(x)=10x^4+4x^2+27x^2+20x+14


Second method is the multiplication rule
p(x)=(x^3+2x+4)*derivative of(2x^2+x+5) + (2x^2+x+5)*derivative of (x^3+2x+4)
p'(x)=(x^3+2x+4)(4x+1) +(2x^2+x+5)(3x^2+2)
p'(x)=10x^4+4x^2+27x^2+20x+14

Answer 4

You could expand it then calculate the derivative of all the terms, but you might as well learn the Chain Rule - it's used a lot.

Chain Rule. Call the first term F(x) = (x^3 + 2x + 4) and the second G(x) = (2x^2 + x + 5).

The chain rule in words is "first term times the derivative of the second term, plus the second term times the derivative of the first term", ie

F * G'(x) + G * F'(x)

so, (x^3 + 2x + 4)*(4x + 1) + (2x^2 + x + 5)*(3x^2 + x)

I'll let you do the expansion.

Edit: oops, always get the names wrong!! Must remember Product Rule not Chain Rule!! duh.

Answer 5

using the f(x) = u.v form,the expansion is
p'(x) = (x3+2x+4)(4x+1) + (3x2+2)(2x2+x+5)
now just expand this exp.
Alternatively,you could expand the original expression and then take derivative.

Answer 6

( x^3 + 2x + 4 )( x^2 + 5x - 3 ) x^5 + 5x^4 - 3x^3 + 2x^3 + 10x^2 - 6x + 4x^2 + 20x - 12 x^5 + 5x^4 - x^3 + 14x^2 + 14x - 12 p'(x) = 5x^4 + 20x^3 - 3x^2 + 28x + 14

Answer 7

p(x) = ( x3+2 x+4 ) ( 2 x2+x+5 )
from the definition
D(f(x)*g(x))=f(x)*D(g(x))+g(x)*D(f(x))
p'(x)=(x^3+2x+4)*(4x+1)+(2x^2+x+5)*(3x^2+2)
p'(x)=(4x^4+8x^2+16x+x^3+2x+4)+(6x^4+3x^3+15x^2+4x^2+2x+10)
p'(x)=10x^4+4x^3+12x^2+20x+14

Answer 8

Since it's multiplication, you can use the "udv + vdu" in getting the derivative.

p(x) = (x3+2x+4) (2x2+x+5)
p'(x) = (x3+2x+4) (4x+1) + (2x2+x+5) (3x2+2)
p'(x) = (4x4+8x+16x+x3+2x+4) + (6x4+3x3+15x2+4x2+2x+10)
p'(x) = 10x4+4x3+27x2+20x+14

Answer 9

mdnif has it right except for the name of the rule. It's the product rule not the chain rule.

Answer 10

(3x^2+2)(4x+1)
(12x^3+8x+3x^2+2)