How do you find the derivative of #f(x) = x^4 - 3/4x^3 - 4x^2 +1#?

1 Answer
Nov 15, 2015

#f'(x)=4x^3-9/4x^2-8x#

Explanation:

Things to know:
- #1#: the derivative of #x^n# is #nx^(n-1)#
- #2#: the derivative of a sum is the sum of the derivatives (i.e., the derivative of #3x+1# is equal to the derivative of #3x# plus the derivative of #1#)
- #3#: scalar constants (constants being multiplied) can be multiplied by the derivative (i.e., the derivative of #9x^2# equals #9# times the derivative of #x^2#)
- #4#: the derivative of a constant is #0#

So, #f'(x)=#the sum of the derivatives of each part. (Rule #2#)

Let's go one at a time:
#d/(dx)[x^4]=4x^3" "("Rule "1)#
#d/(dx)[-3/4x^3]=-9/4x^2" "("Rules "1,3)#
#d/(dx)[-4x^2]=-8x^1=-8x" "("Rules "1,3)#
#d/(dx)[1]=0" "("Rule "4)#

So, we can combine all the derivatives we found the determine that #f'(x)=4x^3-9/4x^2-8xcancel(+0#