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

Nov 15, 2015

$f ' \left(x\right) = 4 {x}^{3} - \frac{9}{4} {x}^{2} - 8 x$

#### Explanation:

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

So, $f ' \left(x\right) =$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