Differentiate y = (e^x + 1)/x^3.

Step-by-step solution

1. 1

   Rewrite the quotient as a product. Moving $x^3$ into the numerator with a negative exponent avoids the bulky quotient-rule denominator:

   $y=(e^x+1)\,x^{-3}$

   The domain excludes $x=0$, where the original function is undefined.

2. 2

   Set up the product rule. With $u=e^x+1$ and $v=x^{-3}$,

   $u'=e^x,\qquad v'=-3x^{-4}$

   Note $u'=e^x$ and not $e^x+1$ — the derivative of the constant $1$ is zero.

3. 3

   Apply $(uv)'=u'v+uv'$.

   $y'=e^x\cdot x^{-3}+(e^x+1)\left(-3x^{-4}\right)=\frac{e^x}{x^3}-\frac{3(e^x+1)}{x^4}$

4. 4

   Put both terms over the common denominator $x^4$. Multiply the first term by $\tfrac{x}{x}$:

   $y'=\frac{x e^x}{x^4}-\frac{3(e^x+1)}{x^4}=\frac{x e^x-3e^x-3}{x^4}$

5. 5

   Factor the exponential to get the cleanest form.

   $y'=\frac{e^x(x-3)-3}{x^4}$

   The stray $-3$ cannot be absorbed into the factor because it comes from differentiating the constant $1$ in the numerator.

6. 6

   Verify numerically at several points. Comparing with a symmetric difference quotient of step $10^{-6}$: at $x=0.6$ both give $-56.8911$; at $x=1.7$ both give $-1.21121$; at $x=-2.2$ both give $-0.152661$ ✓, including a negative $x$ where the odd power $x^3$ matters.