Derivative of e^3x by first principle and chain rule

The derivative of e^3x is 3e^3x. The function e^3x is an exponential function with an exponent 3x. In this note, we will find the derivative of e to the power 3x by the first principle of derivatives and by the chain rule of derivatives.

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Derivative of e^3x using first principle

As we know that the derivative of a function $f(x)$ by first principle is the below limit

$\dfrac{d}{dx}(f(x))$ $=\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h},$

so taking $f(x)=e^{3x}$ in the above equation, the derivative of $e^{3x}$ from first principle is

$\dfrac{d}{dx}(e^{3x})$ $=\lim\limits_{h \to 0} \dfrac{e^{3(x+h)}-e^{3x}}{h}$

$=\lim\limits_{h \to 0} \dfrac{e^{3x+3h}-e^{3x}}{h}$

$=\lim\limits_{h \to 0} \dfrac{e^{3x} \cdot e^{3h}-e^{3x}}{h}$

$=\lim\limits_{h \to 0} \dfrac{e^{3x} (e^{3h}-1)}{h}$

$=e^{3x} \lim\limits_{h \to 0} \dfrac{e^{3h}-1}{3h}$ $\times 3$

Let $t=3h$. Thus $t \to 0$ when $h \to 0.$

So from above, we get $=e^{3x} \lim\limits_{t \to 0} \dfrac{e^{t}-1}{t}$ $\times 3$

$=e^{3x} \times 1 \times 3$ as the limit of $(e^t-1)/t$ is one when t tends to zero.

$=3e^{3x}$

Thus the derivative of e^3x is 3e^3x and this is obtained by the first principle of derivatives.

Now we will find the derivative of e to the power 3x by the chain rule of derivatives.

Derivative of e^3x by Chain Rule

Let $z=3x$. Therefore, we have $dz/dx=3.$ By the chain rule of derivatives, we have

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

$=e^z \cdot 3$ as the derivative of e^z with respect to z is e^z, and $dz/dx=3.$

$=3e^z$

$=3e^{3x}$ as $z=3x$

So the derivative of e^3x is 3e^3x and this is obtained by the chain rule of derivatives.

Question Answer on Derivative of e^3x

Question 1: Find the derivative of e^3.

Answer:

Note that e³ is a constant number as the number e is a constant. We know that the derivative of a constant is zero (see the page on Derivative of a constant is 0). Thus we can say that the derivative of e cube is zero.

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