What is the Derivative of e^{2x}? The derivative of e^{2x }is 2e^{2x}. Here, we will find the derivative of e^{2x} using the derivative of e^{x}. We will use three methods: substitution method, chain rule, and limit definition.
Derivative of e^{2x }Formula
The derivative of e^{2x }is 2e^{2x}. This can be written mathematically as follows: d/dx(e^{2x}) = 2 e^{2x } or (e^{2x})’ = 2 e^{2x}.
What is the derivative of e^{2x}?
At first, we will find the derivative of e^{2x }by the substitution method. This method is known as logarithmic differentiation. The following steps have to be followed in this method.
Step 1: Let
\[y=e^{2x}.\]
Step 2: Taking logarithms on both sides, we get that
\[\log_e y =\log_e e^{2x}\]
\[\Rightarrow \log_e y =2x\log_e e \]
\[\Rightarrow \log_e y =2x [\because \log_a a=1]\]
Step 3: Differentiating both sides of log_{e}y=2x with respect to x, we have
$\dfrac{1}{y} \dfrac{dy}{dx}=\dfrac{d}{dx}(2x)$
$\Rightarrow \dfrac{1}{y} \dfrac{dy}{dx}=2$
$\Rightarrow \dfrac{dy}{dx}=2y$
$\Rightarrow \dfrac{dy}{dx}=2e^{2x}$ $[\because y=e^{2x}]$
∴ The derivative of e^{2x} by the logarithmic differentiation is 2e^{2x}. That is, we have
\[\dfrac{d}{dx}(e^{2x})=2e^{2x}.\]
Derivative of e^{2x} using chain rule
Recall the chain rule of derivatives: $\frac{df}{dx}=\frac{df}{dz} \cdot \frac{dz}{dx}.$ To find the differentiation of e^{2x} by the chain rule, we put
z=2x
Now, $\dfrac{d}{dx}(e^{2x})=\dfrac{d}{dx}(e^z)$
$=\dfrac{d}{dz}(e^z) \cdot \dfrac{dz}{dx}$, by the chain rule.
$=\dfrac{d}{dz}(e^z) \cdot \dfrac{d}{dx}(2x)$
= e^{z} ⋅ 2 as (e^{x})$’$=e^{x}
= 2e^{2x} [∵ z=2x]
∴ the derivative of e^{2x }by chain rule is 2e^{2x}.
Also Read:
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Derivative of e^{2x} from first principle
From first principle of derivatives, we know the derivative of f(x) is given by
$f'(x)=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}$
So taking f(x)=e^{2x}, the derivative of e^{2x} by first principle is
$(e^{2x})’= \lim\limits_{h \to 0} \dfrac{e^{2(x+h)}-e^{2x}}{h}$
$=\lim\limits_{h \to 0} \dfrac{e^{2x+2h}-e^{2x}}{h}$
$=\lim\limits_{h \to 0} \dfrac{e^{2x} \cdot e^{2h}-e^{2x}}{h}$
$=\lim\limits_{h \to 0} \dfrac{e^{2x}(e^{2h}-1)}{h}$
$=e^{2x}\lim\limits_{h \to 0} \Big(\dfrac{e^{2h}-1}{2h} \times 2 \Big)$
$=2e^{2x}\lim\limits_{h \to 0} \dfrac{e^{2h}-1}{2h}$
[Let $t=2h.$ Then $t \to 0$ as $x \to 0$]
$=2e^{2x}\lim\limits_{t \to 0} \dfrac{e^{t}-1}{t}$
$=2e^{2x} \cdot 1$
$=2e^{2x}$
This shows that the differentiation of e^{2x} by first principle is equal to 2e^{2x}.
n-th Derivative of e^{2x}
Differentiating e^{2x} with respect to x for n times, we will get the n^{th} derivative of e to the power x. The first order derivative of e^{2x} is 2e^{2x}. So the second order derivative of e^{2x} is equal to d/dx(2e^{2x}) = 2 d/dx(e^{2x}) = 2×2 e^{2x} = 2^{2} e^{2x}. Now, we will understant the patterns of these derivatives:
- First derivative of e^{2x} is 2e^{2x} = 2^{1} e^{2x}
- Second derivative of e^{2x} is 2^{2 }e^{2x}
- Third derivative of e^{2x} is 2^{3 }e^{2x } and so on.
Observing these patterns carefully, we get the n^{th} derivative of e^{2x} is 2^{n} e^{2x}.
Important Notes on Derivative of e^{2x}:
- The differentiation of e^{x} is e^{x} but d/dx(e^{2x}) = 2e^{2x}.
- The derivative of e^{-2x} is -2e^{-2x}.
- In general, the derivative of e^{mx} is me^{mx} where m is a real number.
Application of Derivative of e^{2x}
From the above we know that
$\dfrac{d}{dx}(e^{2x})$ $=2e^{2x}$ $\cdots (I)$
Using this fact, we can calculate the derivatives of many exponential functions. We will find the derivative of $e^{2x+1}.$
Question 1: Find $\dfrac{d}{dx}(e^{2x+1})$
Solution:
Let z=2x+1.
$\dfrac{d}{dx}(e^{2x+1})$ $=\dfrac{d}{dx}(e^z)$
$=\dfrac{d}{dz}(e^z) \cdot \dfrac{dz}{dx}$ by the chain rule
$=e^z \cdot \dfrac{d}{dx}(2x+1)$ by Equation (I)
$=e^z \cdot 2$
$=2e^{2x+1}$ $[\because z=2x+1]$
Thus, the derivative of e^{2x+1} is 2e^{2x+1}.
Question 2: What is the derivative of e^{2}?
Solution:
Note that the function e^{2 }is a constant function of x. We know that the derivative of a constant function is 0. Thus, the derivative of e square is zero.