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

Question

Emma Aldridge · Accepted Answer

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

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!

Elijah Abram · Answer

undefined To differentiate $ f(x) = (5x + 4x^4)(5 + 3x^2) $ using the product rule:

1. Identify the functions $ u(x) $ and $ v(x) $ where $ u(x) = 5x + 4x^4 $ and $ v(x) = 5 + 3x^2 $.
2. Apply the product rule formula: $ (uv)' = u'v + uv' $.
3. Differentiate $ u(x) $ and $ v(x) $ individually to find $ u'(x) $ and $ v'(x) $.
   - $ u'(x) $ is the derivative of $ u(x) $ with respect to $ x $.
   - $ v'(x) $ is the derivative of $ v(x) $ with respect to $ x $.
4. Substitute the values of $ u(x) $, $ v(x) $, $ u'(x) $, and $ v'(x) $ into the product rule formula.
5. Simplify the expression to get the final result.