Q.Differentiate $a^x$ w.r.t. $x$, where $a$ is a positive constant.

The key idea is that $a^x$ is an exponential function, and its derivative follows from rewriting it using the natural exponential: $a^x = e^{x \log a}$.

Step 1: Write $a^x$ as $e^{x \log a}$.

Step 2: Differentiate using the chain rule. The derivative of $e^{u}$ is $e^{u} \cdot \frac{du}{dx}$, where $u = x \log a$.

Step 3: Since $\log a$ is a constant, $\frac{d}{dx}(x \log a) = \log a$.

Step 4: Multiply: $\frac{d}{dx}(a^x) = e^{x \log a} \cdot \log a = a^x \log a$.

The function $a^x$ is an exponential with a constant base. Unlike $e^x$, whose derivative is itself, $a^x$ has a base that isn't the natural base $e$. The trick is to express any exponential in terms of $e$, because we know exactly how to differentiate $e^{\text{something}}$.

The key identity is $a = e^{\log a}$, so $a^x = (e^{\log a})^x = e^{x \log a}$. Now the exponent is a simple linear function of $x$, and the derivative becomes straightforward.

1. Rewrite the function

   Since $a > 0$, we can write $a^x = e^{x \log a}$. This is valid for all real $x$.

2. Apply the chain rule

   Let $u = x \log a$. Then $a^x = e^u$.

   The chain rule gives:

3. Differentiate the exponent

   $\frac{du}{dx} = \log a$, because $\log a$ is a constant.

4. Combine the results

Method: Differentiating an Exponential Function with a Constant Base

Use this method for any function of the form $a^{x}$ (or more generally $a^{g(x)}$), where the base $a$ is a fixed positive constant and the variable sits only in the exponent.

Steps

Step 1: Rewrite the base-$a$ exponential in terms of the natural base $e$

Every positive $a \neq 1$ can be written as $a = e^{\log a}$, so:

Step 2: Recognize this as a chain-rule composition, $e^u$ with $u = x\log a$

Step 3: Differentiate using the chain rule

since $\log a$ is a constant.

Step 4: Substitute back $e^{x\log a} = a^x$

Applying to this type of problem: treat this as the standard formula to recall directly once derived — $\dfrac{d}{dx}(a^x) = a^x \log a$ — and note it reduces correctly to $e^x$ when $a = e$, since $\log e = 1$.

Common Mistakes

Mistake 1: Applying the power rule instead of the exponential rule

Why it's wrong: In $a^x$, the variable $x$ is in the exponent, not the base — the power rule $\frac{d}{dx}x^n = nx^{n-1}$ only applies when the base is variable and the exponent is a fixed constant, which is the opposite situation here. Writing $x\,a^{x-1}$ mistakenly treats $a^x$ as if it were a power function. Correct approach: recognize that a variable exponent with a fixed base always calls for the exponential-derivative rule, $\frac{d}{dx}a^x = a^x\log a$.

Mistake 2: Forgetting the $\log a$ factor entirely

Why it's wrong: Simply writing $\frac{d}{dx}(a^x) = a^x$ (as if $a$ were $e$) ignores that the chain rule contributes a factor equal to the derivative of the exponent $x\log a$, which is $\log a$ — this factor is only $1$ in the special case $a=e$. Correct approach: always include the $\log a$ multiplier, and only drop it when the base is specifically $e$.