Find y' for y = (x+1)(x-1)
Find $y'$ for $y = (x+1)(x-1)$.
Explanation of Method
Two routes to the same derivative: expand first, or use the product rule.
Step 1: Method 1: expand first
The product is a difference of squares: $y = (x+1)(x-1) = x^{2} - 1$.
Differentiating this polynomial term by term: $y' = 2x - 0 = 2x$.
Step 2: Method 2: the product rule
For $y = uv$ recall $y' = u'v + uv'$. Here $u = x+1$, $u' = 1$, $v = x-1$, $v' = 1$: \[y' = 1\cdot(x-1) + (x+1)\cdot 1 = x - 1 + x + 1 = 2x.\]
Step 3: Verification
\[\text{Both methods agree:}\quad 2x = 2x \;\;\checkmark\]
\[y' = 0 \;\Longleftrightarrow\; x = 0, \text{ the vertex of the parabola } y = x^{2}-1 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = 2x\,}\]
Remark
Remark. Note the common error of writing $(uv)' = u'v'$, which here would give $1\cdot 1 = 1$ instead of $2x$. The product rule has two terms precisely because both factors vary with $x$.