Question

How do you use the alternative definition to find the derivative of $f\left( x \right) = \sqrt {x + 3}$ at $x = 1$?

Answer 1

In the given question, we have been given a linear equation. It is in an expression of basic operations. We have to find the derivative of this linear expression. To do that, we simplify all the functions into the common use, apply the formula of differentiation, plug in the given value, and find the value of the given function. Here, the function has to be derived at a particular constant value.

Formula Used:
We are going to use the formula of differentiation, which is:
$\dfrac{{d\left( {{k^n}} \right)}}{{dx}} = n{k^{n - 1}}$

Complete step by step answer:
The given expression is $f\left( x \right) = \sqrt {x + 3}$.
We are going to use the formula of differentiation, which is:
$\dfrac{{d\left( {{k^n}} \right)}}{{dx}} = n{k^{n - 1}}$
Now, $\sqrt {x + 3} = {\left( {x + 3} \right)^{\dfrac{1}{2}}}$
Here, $k = x + 3$ and $n = \dfrac{1}{2}$, so, $n - 1 = - \dfrac{1}{2}$.
Substituting the values into the formula, we get,
$f'\left( x \right) = \dfrac{1}{2}{\left( {x + 3} \right)^{ - \dfrac{1}{2}}}$
or $f'\left( x \right) = \dfrac{1}{2} \times \dfrac{1}{{\sqrt {\left( {x + 3} \right)} }}$
Putting in $x = 1$, we have,
$f'\left( 1 \right) = \dfrac{1}{2} \times \dfrac{1}{{\sqrt {\left( {1 + 3} \right)} }} = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}$

Hence, the answer is $\dfrac{1}{4}$.

Note:
In the given question, we had been given an algebraic expression. We solved this question by applying the rule of differentiation on the algebraic expression. Care must be taken when the formula is being applied.