Chapter 2.7, Problem 24E

To determine

To find:The derivative of the given function by definition of derivative and state the domain of the function and the domain of its derivative.

Answer to Problem 24E

Derivative of the function is $1$ .

Domain of the function and domain of derivative is set of real number.

Explanation of Solution

Given: $f\left(x\right)=x+\sqrt{x}$

Concept used:

Definition of derivative: $f^{\text{'}}\left(x\right)={\underrightarrow{\lim }}_{h\to 0}\frac{f\left(x+h\right)-f\left(x\right)}{h}$

$f\left(x\right)=x+\sqrt{x}$ be a given function.

It has to find derivative of the function.

  $f\left(x\right)=x+\sqrt{x}$

  $f^{\text{'}}\left(x\right)={\underrightarrow{\lim }}_{h\to 0}\frac{f\left(x+h\right)-f\left(x\right)}{h}$

  $\begin{array}{l}{f}^{\text{'}}\left(x\right)={\underrightarrow{\lim }}_{h\to 0}\frac{\left(x+h\right)+\sqrt{\left(x+h\right)}-x-\sqrt{x}}{h}\\ ={\underrightarrow{\lim }}_{h\to 0}\frac{x+h+\sqrt{\left(x+h\right)}-x-\sqrt{x}}{h}\\ ={\underrightarrow{\lim }}_{h\to 0}\frac{h\left(1+\frac{\sqrt{\left(x+h\right)}}{h}-\frac{\sqrt{x}}{h}\right)}{h}\\ ={\underrightarrow{\lim }}_{h\to 0}\left(1+\frac{\sqrt{\left(x+h\right)}}{h}-\frac{\sqrt{x}}{h}\right)\\ =1\end{array}$

Domain of the function is the set of all possible input of the function, so here domain of the function is set of real numbers.

Domain of the derivative is the set of all points in the domain of function at which function is differentiable.

Hence, domain of derivative and domain of function both are set of real number.