Question

How do you find the derivative of $\sqrt {x + 1}$ using limits?

Answer 1

In this question we have to find the derivative of the given function using limits , which is given by the formula $\dfrac{d}{{dx}}f = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$, now here $f\left( x \right) = \sqrt {x + 1}$ substitute the value in the limit formula now rationalise the numerator and simplify the expression by applying limits , we will get the required result.

Complete step by step solution:
Given expression is $\sqrt {x + 1}$,
Now using the limit formula which is given by $\dfrac{d}{{dx}}f = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$,
Given $f\left( x \right) = \sqrt {x + 1}$,
Now derivate on both sides we get,
$\Rightarrow \dfrac{d}{{dx}}f\left( x \right) = \dfrac{d}{{dx}}\sqrt {x + 1}$,
Using limit formula we get,
Here $f\left( {x + h} \right)$will be $\sqrt {x + h + 1}$ and $f\left( x \right) = \sqrt {x + 1}$, now substitute the values in the formula we get,
$\Rightarrow \dfrac{d}{{dx}}\sqrt {x + 1} = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sqrt {x + h + 1} - \sqrt {x + 1} }}{h}$,
Now rationalize the numerator with its conjugate we get,
$\Rightarrow \dfrac{d}{{dx}}\sqrt {x + 1} = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sqrt {x + h + 1} - \sqrt {x + 1} }}{h} \times \dfrac{{\sqrt {x + h + 1} + \sqrt {x + 1} }}{{\sqrt {x + h + 1} + \sqrt {x + 1} }}$,
Now simplifying we get,
$\Rightarrow \dfrac{d}{{dx}}\sqrt {x + 1} = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {\sqrt {x + h + 1} - \sqrt {x + 1} } \right) \times \left( {\sqrt {x + h + 1} + \sqrt {x + 1} } \right)}}{{h\left( {\sqrt {x + h + 1} + \sqrt {x + 1} } \right)}}$,
Now applying algebraic identity $\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)$,
$\Rightarrow \dfrac{d}{{dx}}\sqrt {x + 1} = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {{{\left( {\sqrt {x + h + 1} } \right)}^2} - {{\left( {\sqrt {x + 1} } \right)}^2}} \right)}}{{h\left( {\sqrt {x + h + 1} + \sqrt {x + 1} } \right)}}$,
Now simplifying we get,
$\Rightarrow \dfrac{d}{{dx}}\sqrt {x + 1} = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {x + h + 1 - x - 1} \right)}}{{h\left( {\sqrt {x + h + 1} + \sqrt {x + 1} } \right)}}$,
Again simplifying we get,
$\Rightarrow \dfrac{d}{{dx}}\sqrt {x + 1} = \mathop {\lim }\limits_{h \to 0} \dfrac{h}{{h\left( {\sqrt {x + h + 1} + \sqrt {x + 1} } \right)}}$,
Now eliminating the like terms we get,
$\Rightarrow \dfrac{d}{{dx}}\sqrt {x + 1} = \mathop {\lim }\limits_{h \to 0} \dfrac{1}{{\left( {\sqrt {x + h + 1} + \sqrt {x + 1} } \right)}}$,
Now applying the limits directly we get,
$\Rightarrow \dfrac{d}{{dx}}\sqrt {x + 1} = \dfrac{1}{{\left( {\sqrt {x + 0 + 1} + \sqrt {x + 1} } \right)}}$,
Now simplifying we get,
$\Rightarrow \dfrac{d}{{dx}}\sqrt {x + 1} = \dfrac{1}{{\left( {\sqrt {x + 1} + \sqrt {x + 1} } \right)}}$,
Further simplifying we get,
$\Rightarrow \dfrac{d}{{dx}}\sqrt {x + 1} = \dfrac{1}{{2\sqrt {x + 1} }}$,
So, the derivative of the function is $\dfrac{1}{{2\sqrt {x + 1} }}$.

Final Answer:
$\therefore$The derivative of the given function $\sqrt {x + 1}$will be equal to $\dfrac{1}{{2\sqrt {x + 1} }}$.

Note:
Because differential calculus is based on the definition of the derivative, and the definition of the derivative involves a limit, there is a sense in which all of calculus rests on limits. In addition, the limit involved in the limit definition of the derivative is one that always generates an indeterminate form of $\dfrac{0}{0}$. If $f$is a differentiable function for which ${f^{'}}\left( x \right)$ exists, then when we consider:
$\dfrac{d}{{dx}}{f^{'}}\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$,
If the limit exists, $f$ is said to be differentiable at $x$, otherwise $f$ is non-differentiable at $x$.