Question

How do you find the derivative of: $ f(x) = \sqrt {x + 1} $ using the limit definition?

Answer 1

Hint: In order to determine the derivative of $ f(x) $ with respect to variable x , use the formula of limit definition \[f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\] . To find the $ f\left( {x + h} \right) $ replace all the $ x $ with $ x + h $ in the original function . Now to calculate the limit multiply and divide the limit with \[\sqrt {x + h + 1} + \sqrt {x + 1} \] and use the identity $ \left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2} $ to simplify the numerator and now put the limit $ h = 0 $ to obtain your required derivative.

Complete step-by-step answer:
We are given a function in variable $ x $ i.e.
 $ f(x) = \sqrt {x + 1} $
Let’s look into the formula of limit definition to find the derivative of $ f(x) $ with respect to x
 \[\dfrac{d}{{dx}}\left( {f(x)} \right) = f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\]
 $ f(x) = \sqrt {x + 1} $ --(1)
To find the $ f\left( {x + h} \right) $ replace all the $ x $ with $ x + h $ in the equation (1), we get
 $ f(x + h) = \sqrt {x + h + 1} $
Now putting $ f\left( h \right) $ and $ f\left( {x + h} \right) $ in the formula ,
 \[
  f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h} \\
  f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sqrt {x + h + 1} - \sqrt {x + 1} }}{h} \\
 \]
As you can see this is the limit problem, so if we directly put the limit $ h \to 0 $ the result will have a denominator equal to zero which is completely not acceptable as it gives the result as infinity.
So to avoid this thing , multiply divide the equation \[\sqrt {x + h + 1} + \sqrt {x + 1} \] , we get
 \[f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sqrt {x + h + 1} - \sqrt {x + 1} }}{h} \times \dfrac{{\sqrt {x + h + 1} + \sqrt {x + 1} }}{{\sqrt {x + h + 1} + \sqrt {x + 1} }}\]
Now applying the identity of $ \left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2} $ by considering $ a $ as \[\sqrt {x + h + 1} \] and $ b $ as \[\sqrt {x + 1} \] in the numerator
 \[
  f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\left( {\sqrt {x + h + 1} } \right)}^2} - {{\left( {\sqrt {x + 1} } \right)}^2}}}{{h\left( {\sqrt {x + h + 1} + \sqrt {x + 1} } \right)}} \\
  f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {x + h + 1} \right) - \left( {x + 1} \right)}}{{h\left( {\sqrt {x + h + 1} + \sqrt {x + 1} } \right)}} \\
  f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{x + h + 1 - x - 1}}{{h\left( {\sqrt {x + h + 1} + \sqrt {x + 1} } \right)}} \\
  f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{h}{{h\left( {\sqrt {x + h + 1} + \sqrt {x + 1} } \right)}} \\
  f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{1}{{\left( {\sqrt {x + h + 1} + \sqrt {x + 1} } \right)}} \;
 \]
Now putting the limit $ h = 0 $ , we get
 \[
  f'\left( x \right) = \dfrac{1}{{\left( {\sqrt {x + 0 + 1} + \sqrt {x + 1} } \right)}} \\
  f'\left( x \right) = \dfrac{1}{{\left( {\sqrt {x + 1} + \sqrt {x + 1} } \right)}} \\
  f'\left( x \right) = \dfrac{1}{{2\sqrt {x + 1} }} \;
 \]
Therefore the derivative of $ f(x) = \sqrt {x + 1} $ using the limit definition is equal to \[\dfrac{1}{{2\sqrt {x + 1} }}\] .
So, the correct answer is “ \[\dfrac{1}{{2\sqrt {x + 1} }}\] ”.

Note: 1. Don’t forget to cross-check your answer.
2.After putting the Limit the result should never contain zero in the denominator. If it is contained, apply some operation to modify the result and then after put the limit.
3. Derivative is the inverse of integration .