Find the n -th derivative of y = a^x, a>0
Find the $n$-th derivative of $y = a^{x}$, $a>0$.
Explanation of Method
Every step multiplies by the same logarithm.
Step 1: The basic rule
Writing $a^{x} = e^{x\ln a}$ and applying the chain rule gives $\bigl(a^{x}\bigr)' = a^{x}\ln a$. Repeating: \[y' = a^{x}\ln a, \qquad y'' = a^{x}\bigl(\ln a\bigr)^{2}, \qquad y''' = a^{x}\bigl(\ln a\bigr)^{3}.\]
Step 2: Induction
Claim: $y^{(k)} = a^{x}\bigl(\ln a\bigr)^{k}$. The base case is the first line. Assuming it for $k$: $y^{(k+1)} = \bigl(\ln a\bigr)^{k}\cdot a^{x}\ln a = a^{x}\bigl(\ln a\bigr)^{k+1}$.
Step 3: Verification
\[a = e:\ \ln e = 1 \;\Longrightarrow\; y^{(n)} = e^{x}, \text{ Problem 286} \;\;\checkmark\]
\[a^{x} = \sum_{n=0}^{\infty}\frac{\bigl(\ln a\bigr)^{n}}{n!}x^{n}, \quad\text{coefficient } \frac{y^{(n)}(0)}{n!} = \frac{\bigl(\ln a\bigr)^{n}}{n!} \;\;\checkmark\]
\[a = 2,\ x = 0,\ n = 3:\ y'''(0) = \bigl(\ln2\bigr)^{3} \approx 0.3330 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y^{(n)} = a^{x}\bigl(\ln a\bigr)^{n}\,}\]
Remark
Remark. The behaviour splits at $a = e$. For $a>e$ we have $\ln a>1$ and the derivatives grow without bound; for $1\lt a\lt e$ they shrink; for $a\lt 1$ the logarithm is negative and the signs alternate. The case $a = 1$ is degenerate: $\ln1 = 0$, $y\equiv1$, and every derivative vanishes.