Question

Find the derivative $\dfrac{{d{{\left( {1 + {x^2}} \right)}^{\dfrac{1}{2}}}}}{{dx}}$.

Answer 1

Hint: n the given problem, we are required to differentiate ${\left( {1 + {x^2}} \right)^{\dfrac{1}{2}}}$ with respect to x. Since, ${\left( {1 + {x^2}} \right)^{\dfrac{1}{2}}}$ is a composite function, so we will have to apply chain rule of differentiation in the process of differentiating ${\left( {1 + {x^2}} \right)^{\dfrac{1}{2}}}$ . So, differentiation of ${\left( {1 + {x^2}} \right)^{\dfrac{1}{2}}}$ with respect to x will be done layer by layer using the chain rule of differentiation. Also the power rule of differentiation must be remembered while solving the given problem.

Complete step by step answer:
Now, $\dfrac{{d{{\left( {1 + {x^2}} \right)}^{\dfrac{1}{2}}}}}{{dx}}$
Taking the power outside the bracket in order to apply chain rule of differentiation.
$\Rightarrow \dfrac{d}{{dx}}{\left[ {1 + {x^2}} \right]^{\dfrac{1}{2}}}$
Now, Let us assume $u = \left[ {1 + {x^2}} \right]$. So substituting $\left[ {1 + {x^2}} \right]$ as $u$, we get,
\[\Rightarrow \dfrac{d}{{dx}}{\left[ u \right]^{\dfrac{1}{2}}}\]
Now, using the power rule of differentiation $\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{\left( {n - 1} \right)}}$, we get,
\[\Rightarrow \left( {\dfrac{1}{2}{u^{\left( {\dfrac{1}{2} - 1} \right)}}} \right)\dfrac{{du}}{{dx}}\]

Simplifying the expression further, we get,
\[\Rightarrow \left( {\dfrac{1}{2}{u^{ - \dfrac{1}{2}}}} \right)\dfrac{{du}}{{dx}}\]
Now, putting back $u$as $\left[ {1 + {x^2}} \right]$, we get,
\[\Rightarrow \left( {\dfrac{1}{2}{{\left[ {1 + {x^2}} \right]}^{ - \dfrac{1}{2}}}} \right)\dfrac{{d\left[ {1 + {x^2}} \right]}}{{dx}}\]
Simplifying the expression, we get,
\[\Rightarrow \left( {\dfrac{1}{{2\sqrt {1 + {x^2}} }}} \right)\dfrac{{d\left[ {1 + {x^2}} \right]}}{{dx}}\]
Now, we know that the derivative of a constant is zero. Also, using the power rule of differentiation again, we get,
\[\Rightarrow \left( {\dfrac{1}{{2\sqrt {1 + {x^2}} }}} \right)\left( {2x} \right)\]
Cancelling the common factors in numerator and denominator, we get,
\[ \therefore \left( {\dfrac{x}{{\sqrt {1 + {x^2}} }}} \right)\]

So, the value of $\dfrac{{d{{\left( {1 + {x^2}} \right)}^{\dfrac{1}{2}}}}}{{dx}}$ is \[\left( {\dfrac{x}{{\sqrt {1 + {x^2}} }}} \right)\].

Note: The given problem may also be solved using the first principle of differentiation. The derivatives of basic trigonometric functions must be learned by heart in order to find derivatives of complex composite functions using chain rule of differentiation.The chain rule of differentiation involves differentiating a composite by introducing new unknowns to ease the process and examine the behaviour of function layer by layer. We should take care while doing the calculative steps as it changes our final answer.