Find y' for y = mx^n - nx^m, where m and n are constants
Find $y'$ for $y = mx^{n} - nx^{m}$, where $m$ and $n$ are constants.
Explanation of Method
Differentiate a symmetric combination of two power terms.
Step 1: Differentiate term by term
In each term the constant factor stays put while the power rule acts on the power of $x$: $y' = m\cdot n x^{n-1} - n\cdot m x^{m-1}$.
Step 2: Factor
Both terms now carry the same constant $mn$: $y' = mn\,x^{n-1} - mn\,x^{m-1} = mn\left(x^{n-1} - x^{m-1}\right)$.
Step 3: Verification
\[m = n:\quad y = m x^{m} - m x^{m} \equiv 0 \;\Longrightarrow\; y' = 0, \text{ and the formula gives } m^{2}\bigl(x^{m-1}-x^{m-1}\bigr) = 0 \;\;\checkmark\]
\[m = 2,\ n = 3:\quad y = 2x^{3}-3x^{2} \;\Longrightarrow\; y' = 6x^{2}-6x = 6\bigl(x^{2}-x\bigr) \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = mn\left(x^{\,n-1} - x^{\,m-1}\right)\,}\]
Remark
Remark. The derivative vanishes whenever $x^{n-1} = x^{m-1}$, which for $m \neq n$ happens at $x = 1$ (and at $x = 0$ when both exponents are positive). The symmetry of the expression under swapping $m \leftrightarrow n$ only changes the overall sign.