Using mathematical induction,prove that $\frac{d}{dx}(x^n) = nx^{n-1}$ for all positive integers $n$.

(N/A) To prove: $P(n): \frac{d}{dx}(x^n) = nx^{n-1}$ for all positive integers $n$.
Step $1$: For $n=1$,
$P(1): \frac{d}{dx}(x) = 1 = 1 \cdot x^{1-1}$.
Thus,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some positive integer $k$.
That is,$P(k): \frac{d}{dx}(x^k) = kx^{k-1}$.
Step $3$: We need to prove $P(k+1)$ is true.
Consider $\frac{d}{dx}(x^{k+1}) = \frac{d}{dx}(x \cdot x^k)$.
Using the product rule: $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$.
$= x^k \cdot \frac{d}{dx}(x) + x \cdot \frac{d}{dx}(x^k)$
$= x^k \cdot 1 + x \cdot (kx^{k-1})$
$= x^k + kx^k$
$= (k+1)x^k$
$= (k+1)x^{(k+1)-1}$.
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Therefore,by the principle of mathematical induction,the statement $P(n)$ is true for every positive integer $n$.