The derivative of 3^{x} is equal to 3^{x} ln 3. Here ln 3 denotes the natural logarithm of 3, that is, ln 3 = log_{e} 3. In this post, we will find the derivative of 3 to the power x by the first principle of derivatives.

We first recall the first principle of derivatives. The derivative of a function f(x) by first principle is given by the limit below:

d/dx (f(x)) = lim_{h→0} $\dfrac{f(x+h)-f(x)}{h}$ …(I)

## Derivative of 3^{x} from First Principle

To find the derivative of 3 to the x, we put f(x)=3^{x} in the above formula **(I)**. Then by the first principle, the derivative of 3^{x} is given by

$\dfrac{d}{dx}$(3^{x}) = lim_{h→0} $\dfrac{3^{x+h}-3^x}{h}$

= lim_{h→0} $\dfrac{3^x \cdot 3^h -3^x}{h}$, by the indices rule: a^{m+n} = a^{m} ⋅ a^{n}

= lim_{h→0} $\dfrac{3^x (3^h -1)}{h}$

= 3^{x} lim_{h→0} $\dfrac{3^h -1}{h}$

= 3^{x} log_{e}3, here we have used the limit formula: lim_{t→0} (a^{t}-1)/t = log_{e} a.

= 3^{x} ln 3

Thus, the derivative of 3^{x} is equal to 3^{x} ln 3 and this is obtained by the first principle of derivatives.

**Note**: From the derivative formula of a^{x}, we can obtain the derivative of 3^{x}. We know that $\dfrac{d}{dx}(a^x)=a^x \ln a$. Thus, the derivative of 3^{x} is equal to $\dfrac{d}{dx}(3^x)$ = 3^{x} ln 3.

**Question:** Find the derivative of 3^{2x}. That is, find $\dfrac{d}{dx}(3^{2x})$.

*Answer: *

Let z=2x. Then we have $\dfrac{dz}{dx}=2$. We will use the chain rule of derivatives. Thus, the derivative of 3^{2x} is given by

$\dfrac{d}{dx}(3^{2x})$ $=\dfrac{d}{dz}(3^z) \cdot \dfrac{dz}{dx}$

= 3^{z} ln3 × 2, as we have from the above that $\dfrac{d}{dx}$(3^{x})=3^{x} ln 3.

= 2 ⋅ 3^{2x} ln 3, as we know z=2x.

Thus, the derivative of 3^{2x} is equal to 2⋅ 3^{2x} ln 3.

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## FAQs

**Q1: What is the derivative of 3 ^{x}?**

**Answer:** The derivative of 3^{x} is 3^{x} log_{e}3.