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

Question

Cameron Alexander · Accepted Answer

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

Ezekiel Alexander · Answer

undefined To find the derivative of $ f(x) = x^4 - \frac{3}{4}x^3 - 4x^2 + 1 $, you apply the power rule for derivatives. The power rule states that if $ f(x) = x^n $, then $ f'(x) = nx^{n-1} $. Applying this rule to each term:

- The derivative of $ x^4 $ is $ 4x^{4-1} = 4x^3 $.
- The derivative of $ \frac{3}{4}x^3 $ is $ \frac{3}{4} \times 3x^{3-1} = \frac{9}{4}x^2 $.
- The derivative of $ 4x^2 $ is $ 4 \times 2x^{2-1} = 8x $.
- The derivative of $ 1 $ is $ 0 $ because it's a constant term.

So, putting it all together, the derivative of $ f(x) $ is $ f'(x) = 4x^3 - \frac{9}{4}x^2 - 8x $.