Question

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

Answer 1

Hint: 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 \[x\].