Differentiate the following by first principle: ${{e}^{3x}}$.
Answer
278.1k+ views
Hint: For solving this question you should know about the differentiation of ${{e}^{x}}$ type functions. As we know that the differentiation of exponential functions is the same as that and the power of that will also be differentiated, so we will perform both differentiation and then finally solve it.
Complete step-by-step solution:
According to our question, we have to differentiate ${{e}^{3x}}$ by first principle. Derivative by first principle refers to using algebra to find a general expression for the slope of a curve. It is also known as the delta method. The derivative is a measure of the instantaneous rate of change, which is equal to:
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$
This expression is the foundation for the rest of the differential calculus, every rule, identity and fact follows this. The general notion of rate of change of a quantity y with respect to x is the change in y divided by the change in x about the point a. This describes the average rate of change and can be expressed as:
$\dfrac{f\left( x \right)-f\left( a \right)}{x-a}$
Let us find the derivative of ${{e}^{3x}}$ with the first principle method. Therefore suppose,
$f\left( x \right)={{e}^{3x}}$
On using the first principle,
$\begin{align}
& f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h} \\
& \Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{3\left( x+h \right)}}-{{e}^{3x}}}{h} \\
& \Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{3x}}\left( {{e}^{3h}}-1 \right)3}{3h} \\
\end{align}$
By using $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{x}}-1}{x}=1$, we get,
$f'\left( x \right)=3.{{e}^{3x}}$
So, we get the answer as $3{{e}^{3x}}$.
Note: While solving these types of questions you have to keep in mind that you should solve this from the first principle method, because it is given to do so. Otherwise we can solve it directly also without using the first principle method. You should be careful about calculations too.
Complete step-by-step solution:
According to our question, we have to differentiate ${{e}^{3x}}$ by first principle. Derivative by first principle refers to using algebra to find a general expression for the slope of a curve. It is also known as the delta method. The derivative is a measure of the instantaneous rate of change, which is equal to:
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$
This expression is the foundation for the rest of the differential calculus, every rule, identity and fact follows this. The general notion of rate of change of a quantity y with respect to x is the change in y divided by the change in x about the point a. This describes the average rate of change and can be expressed as:
$\dfrac{f\left( x \right)-f\left( a \right)}{x-a}$
Let us find the derivative of ${{e}^{3x}}$ with the first principle method. Therefore suppose,
$f\left( x \right)={{e}^{3x}}$
On using the first principle,
$\begin{align}
& f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h} \\
& \Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{3\left( x+h \right)}}-{{e}^{3x}}}{h} \\
& \Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{3x}}\left( {{e}^{3h}}-1 \right)3}{3h} \\
\end{align}$
By using $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{x}}-1}{x}=1$, we get,
$f'\left( x \right)=3.{{e}^{3x}}$
So, we get the answer as $3{{e}^{3x}}$.
Note: While solving these types of questions you have to keep in mind that you should solve this from the first principle method, because it is given to do so. Otherwise we can solve it directly also without using the first principle method. You should be careful about calculations too.
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