How do you differentiate #f(x)= (5x+4x^4)(5+3x^2) # using the product rule?

1 Answer
Aviv S.
Mar 23, 2018

The derivative of #f(x)# is #72x^5+80x^3+45x^2+25#.

Explanation:

Here's the product rule:

#d/dx(fg)=fd/dx(g)+gd/dx(f)#

An easy way to remember this is to say "Left d-Right, Right d-Left", meaning the product is the left function times the derivative of the right, plus the right function times the derivative of the left.

Here's the product rule applied to our problem:

#color(white)=d/dx(5x+4x^4)(5+3x^2)#

#=(5x+4x^4)*d/dx(5+3x^2)+(5+3x^2)*d/dx(5x+4x^4)#

#=(5x+4x^4)\*(0+6x)+(5+3x^2)*(5+16x^3)#

#=(5x+4x^4)\*6x+(5+3x^2)*(5+16x^3)#

#=30x^2+24x^5+(5+3x^2)*(5+16x^3)#

#=30x^2+24x^5+25+80x^3+15x^2+48x^5#

#=24x^5+25+80x^3+45x^2+48x^5#

#=72x^5+25+80x^3+45x^2#

#=72x^5+80x^3+45x^2+25#

That's the derivative. Hope this helped!

Related questions
See all questions in Product Rule
Impact of this question
1705 views around the world
You can reuse this answer
Creative Commons License