Differentiate y = 8x⁵ - 3x to the minus 4, plus 4 over x, minus 9.

Step-by-step solution

1. 1

   Put all four terms in the form $cx^n$. Only the third term needs work:

   $\frac4x=4x^{-1}$

   so $y=8x^5-3x^{-4}+4x^{-1}-9$. Doing this conversion up front means one single rule handles the whole function.

2. 2

   Differentiate the leading term.

   $\frac{d}{dx}\left(8x^5\right)=8\cdot 5x^4=40x^4$

   The exponent $5$ becomes a factor and drops to $4$.

3. 3

   Differentiate $-3x^{-4}$.

   $\frac{d}{dx}\left(-3x^{-4}\right)=-3\cdot(-4)x^{-4-1}=12x^{-5}$

   Subtracting one from $-4$ gives $-5$: the exponent becomes more negative, which is the step most often reversed by mistake.

4. 4

   Differentiate $4x^{-1}$, and note the constant dies.

   $\frac{d}{dx}\left(4x^{-1}\right)=-4x^{-2},\qquad \frac{d}{dx}(-9)=0$

   The derivative of any constant is zero because a horizontal line has slope zero.

5. 5

   Assemble the answer with positive exponents.

   $\frac{dy}{dx}=40x^4+\frac{12}{x^5}-\frac{4}{x^2}$

6. 6

   Check at $x=1$ and note the domain. The formula gives $40+12-4=48$, matching a numerical difference quotient of the original function at $x=1$. Both $y$ and $y'$ are undefined at $x=0$, so the result holds for $x\neq 0$.