Question

Find the derivative of the function $f\left( x \right)={{e}^{\sqrt{x}}}$w.r.t. x from the first principle.

Answer 1

Hint: First, before proceeding for this, we must know the formula for the first principle to calculate the derivative of the function as $\dfrac{d}{dx}f\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$. Then, by substituting the value of function f(x) and f(x+h) in the first principle formula and then by using the formula of the limits and multiplying the dividing the above expression by $\sqrt{x+h}-\sqrt{x}$, we proceed for the solution. Then, by using the rationalisation method, we get the final result after substitution of limit of h.

Complete step by step answer:
In this question, we are supposed to find the derivative of the function $f\left( x \right)={{e}^{\sqrt{x}}}$with respect to x by using the first principle.
So, before proceeding for this, we must know the formula for the first principle to calculate the derivative of the function as:
$\dfrac{d}{dx}f\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$
Now, we need to get the value of f(x+h) by replacing x by (x+h) in the given function, we get:
$f\left( x+h \right)={{e}^{\sqrt{x+h}}}$
Then, by substituting the value of function f(x) and f(x+h) in the first principle formula, we get:
$\dfrac{d}{dx}f\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{\sqrt{x+h}}}-{{e}^{\sqrt{x}}}}{h}$
Now, by solving the above expression, we get:
$\dfrac{d}{dx}f\left( x \right)={{e}^{\sqrt{x}}}\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{\sqrt{x+h}-\sqrt{x}}}-1}{h}$
Now, by using the formula of the limits and multiplying the dividing the above expression by $\sqrt{x+h}-\sqrt{x}$, we get:
$\dfrac{d}{dx}f\left( x \right)={{e}^{\sqrt{x}}}\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{\sqrt{x+h}-\sqrt{x}}}-1}{h}\times \dfrac{\sqrt{x+h}-\sqrt{x}}{\sqrt{x+h}-\sqrt{x}}$
Then, by solving the above expression using the formulas for limit of exponential function as $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{x-1}}}{x}=1$, we get:
$\begin{align}
  & \dfrac{d}{dx}f\left( x \right)={{e}^{\sqrt{x}}}\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{\sqrt{x+h}-\sqrt{x}}}-1}{\sqrt{x+h}-\sqrt{x}}\times \dfrac{\sqrt{x+h}-\sqrt{x}}{h} \\
 & \Rightarrow \dfrac{d}{dx}f\left( x \right)={{e}^{\sqrt{x}}}\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{x+h}-\sqrt{x}}{h} \\
\end{align}$
Then, by using the rationalisation method, we will proceed further as:
$\dfrac{d}{dx}f\left( x \right)={{e}^{\sqrt{x}}}\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{x+h}-\sqrt{x}}{h}\times \dfrac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}$
Then, by using the identity as $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we get:
$\begin{align}
  & \dfrac{d}{dx}f\left( x \right)={{e}^{\sqrt{x}}}\underset{h\to 0}{\mathop{\lim }}\,\dfrac{x+h-x}{h}\times \dfrac{1}{\sqrt{x+h}+\sqrt{x}} \\
 & \Rightarrow \dfrac{d}{dx}f\left( x \right)={{e}^{\sqrt{x}}}\underset{h\to 0}{\mathop{\lim }}\,\dfrac{h}{h}\times \dfrac{1}{\sqrt{x+h}+\sqrt{x}} \\
 & \Rightarrow \dfrac{d}{dx}f\left( x \right)={{e}^{\sqrt{x}}}\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{\sqrt{x+h}+\sqrt{x}} \\
\end{align}$
Then, by substituting the value of h as 0 given in the limit, we get:
$\begin{align}
  & \dfrac{d}{dx}f\left( x \right)={{e}^{\sqrt{x}}}\dfrac{1}{\sqrt{x+0}+\sqrt{x}} \\
 & \dfrac{d}{dx}f\left( x \right)={{e}^{\sqrt{x}}}\dfrac{1}{2\sqrt{x}} \\
\end{align}$

Hence, we get the derivative of the function $f\left( x \right)={{e}^{\sqrt{x}}}$using first principle as ${{e}^{\sqrt{x}}}\dfrac{1}{2\sqrt{x}}$.

Note: Now, to solve these types of questions we need to know some of the basic formulas of limits to get the answer accurately and easily. So, the required formula for the above question is as:
$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{x-1}}}{x}=1$. Moreover, there is another way to find the derivative of the function which is chain rule but here the first principle method is asked.