Question

How do you find the second derivation of ${x^2} + {y^2} = 1$?

Answer 1

Hint: According to the question we have to find the second derivation of ${x^2} + {y^2} = 1$. So, first of all we have to find the first derivative of the given expression with respect to $x$ with the help of the formula given below.

Formula used: $\dfrac{d}{{da}}\left( {{a^2}} \right) = 2a$, $\dfrac{d}{{da}}\left( {{b^2}} \right) = 2b\dfrac{{db}}{{da}}..............................(A)$
Now, we have to find the second derivative to the expression obtained from the first derivative with the help of the formula given below.
$
   \Rightarrow \dfrac{d}{{da}}\left( {a.b} \right) = a.\dfrac{{db}}{{da}} + b.\dfrac{d}{{da}}\left( a \right) \\
   \Rightarrow \dfrac{d}{{da}}\left( {a.b} \right) = a.\dfrac{{db}}{{da}} + b.a............................(B)
 $

Complete step-by-step solution:
Step 1: First of all we have to find the first derivative of the given expression ${x^2} + {y^2} = 1$ with respect to $x$
 $
\Rightarrow \dfrac{d}{{dx}}\left( {{x^2} + {y^2} = 1} \right) \\
\Rightarrow \dfrac{d}{{dx}}\left( {{x^2}} \right) + \dfrac{d}{{dx}}\left( {{y^2}} \right) = \dfrac{d}{{dx}}\left( 1 \right)
 $
Now, we have to use the formula (A) as mentioned in the solution hint and as we know that $\dfrac{d}{{dx}}$ of any constant number is 0.
$
   \Rightarrow 2x + 2y\dfrac{{dy}}{{dx}} = 0.....................(1) \\
   \Rightarrow 2y\dfrac{{dy}}{{dx}} = - 2x \\
   \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{x}{y}
 $
Step 2: Now, we have to find the second derivative to the expression (1) obtained in the solution step 1 and using the formula (B) as mentioned in the solution hint.
\[
\Rightarrow \dfrac{d}{{dx}}(2x) + \dfrac{d}{{dx}}\left( {2y\dfrac{{dy}}{{dx}}} \right) = 0 \\
\Rightarrow 2 + 2\left[ {y.\dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) + \dfrac{{dy}}{{dx}}.\dfrac{d}{{dx}}\left( y \right)} \right] = 0 \\
   \Rightarrow 2 + 2\left[ {y.\dfrac{{{d^2}y}}{{d{x^2}}} + {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}} \right] = 0
 \]
Step 3: Now, we have to put the value of $\dfrac{{dy}}{{dx}}$ as obtained in the solution step 1 in the expression obtained in the solution step 2.
\[
   \Rightarrow 2 + 2\left[ {y.\dfrac{{{d^2}y}}{{d{x^2}}} + {{\left( {\dfrac{{ - x}}{y}} \right)}^2}} \right] = 0 \\
   \Rightarrow 2 + 2y.\dfrac{{{d^2}y}}{{d{x^2}}} + \dfrac{{2{x^2}}}{{{y^2}}} = 0 \\
   \Rightarrow 2y.\dfrac{{{d^2}y}}{{d{x^2}}} = - 2 - \dfrac{{2{x^2}}}{{{y^2}}} \\
   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{ - 2}}{{2y}} - \dfrac{{ - 2{x^2}/{y^2}}}{{2y}} \\
   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{ - 1}}{y} - \dfrac{{ - {x^2}}}{{{y^3}}}
 \]

Hence, the second derivative of the given expression ${x^2} + {y^2} = 1$ is \[\dfrac{{ - 1}}{y} - \dfrac{{ - {x^2}}}{{{y^3}}}\].

Note: It is necessary to find the solution first we have to determine the first derivative and then we have to determine the second derivative.
To determine the second derivative of the expression after the first derivation we have to apply the derivation again for the obtained expression.