# Derivative of 3^x by First Principle

The derivative of 3x is equal to 3x ln 3. Here ln 3 denotes the natural logarithm of 3, that is, ln 3 = loge 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)) = limh→0 $\dfrac{f(x+h)-f(x)}{h}$ …(I)

## Derivative of 3x from First Principle

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

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

= limh→0 $\dfrac{3^x \cdot 3^h -3^x}{h}$, by the indices rule: am+n = am ⋅ an

= limh→0 $\dfrac{3^x (3^h -1)}{h}$

= 3x limh→0 $\dfrac{3^h -1}{h}$

= 3x loge3, here we have used the limit formula: limt→0 (at-1)/t = loge a.

= 3x ln 3

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

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

Question: Find the derivative of 32x. That is, find $\dfrac{d}{dx}(3^{2x})$.

Let z=2x. Then we have $\dfrac{dz}{dx}=2$. We will use the chain rule of derivatives. Thus, the derivative of 32x is given by

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

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

= 2 ⋅ 32x ln 3, as we know z=2x.

Thus, the derivative of 32x is equal to 2⋅ 32x ln 3.

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## FAQs

Q1: What is the derivative of 3x?

Answer: The derivative of 3x is 3x loge3.