Question

If\[{x^2} + {y^2} = 1\] then \[\left( {{y^\prime } = \frac{{dy}}{{dx}},{y^{\prime \prime }} = \frac{{{d^2}y}}{{d{x^2}}}} \right)\]
A. \[y{y^{\prime \prime }} - 2{\left( {{y^\prime }} \right)^2} + 1 = 0\]
B. \[y{y^{\prime \prime }} + {\left( {{y^\prime }} \right)^2} + 1 = 0\]
C. \[y{y^{\prime \prime }} - {\left( {{y^\prime }} \right)^2} - 1 = 0\]
D. \[y{y^{\prime \prime }} + 2{\left( {{y^\prime }} \right)^2} + 1 = 0\]

Answer 1

Hint :
To answer this question, we will utilize the chain rule of derivatives and the formulas \[y' = \frac{{dy}}{{dx}}{\rm{\& }}y'' = \frac{{{d^2}y}}{{d{x^2}}}\] to simultaneously compute the derivatives on both sides. First, differentiate the supplied term with respect to x on both sides, then take the two common terms from the differential to simplify it and last, differentiate the differential once again with respect to x on both sides to determine which of the given options the correct one is.
Formula use:
\[y = {x^n}\]
Differentiate with respect to x
\[\frac{{dy}}{{dx}} = n{x^{n-1}}\]
\[y' = \frac{{dy}}{{dx}}{\rm{\& }} y'' = \frac{{{d^2}y}}{{d{x^2}}}\]
Complete step-by-step solution:
We have been given the equation in the question \[{x^2} + {y^2} = 1\] and we have to find \[\left( {{y^\prime } = \frac{{dy}}{{dx}},{y^{\prime \prime }} = \frac{{{d^2}y}}{{d{x^2}}}} \right)\]
On differentiating either side of the above equation with respect to\[x\], we get
Therefore,
\[\frac{d}{{dx}}\left( {{x^2} + {y^2}} \right) = \frac{d}{{dx}} \times 1\]
On multiplying the above equation with \[\frac{d}{{dx}}\] to the terms inside the parentheses, we obtain
\[ \Rightarrow \frac{d}{{dx}} \times {x^2} + \frac{d}{{dx}} \times {y^2} = 0\]
We have to differentiate the above equation again, we get
\[ \Rightarrow 2x + 2y \times \frac{{dy}}{{dx}} = 0\]
On taking \[2\] as common from the above equation, we have
\[ \Rightarrow 2\left( {x + y \times \frac{{dy}}{{dx}}} \right) = 0\]
Now, we have to divide either side of the above equation by \[2\], we obtain
\[ \Rightarrow x + y \times \frac{{dy}}{{dx}} = 0\]
After doing the first derivative, the above equation is what we obtained.
On differentiating the obtained equation, one more time we get
\[\frac{{dx}}{{dx}} + \frac{d}{{dx}}\left( {y \times \frac{{dy}}{{dx}}} \right) = 0\]
Now, we have to expand the above equation, we get
\[ \Rightarrow 1 + y \times \frac{{d\frac{{dy}}{{dx}}}}{{dx}} + \frac{{dy}}{{dx}} \times \frac{{dy}}{{dx}} = 0\]
On simplifying the above equation, we get
\[ \Rightarrow 1 + y \times \frac{{{d^2}y}}{{d{x^2}}} + {\left( {\frac{{dy}}{{dx}}} \right)^2} = 0\]
Now, we have to rewrite the above equation as
\[ \Rightarrow y\left( {{y^{\prime \prime }}} \right) + {\left( {{y^\prime }} \right)^2} + 1 = 0\]
Therefore, if \[{x^2} + {y^2} = 1\] then \[\left( {{y^\prime } = \frac{{dy}}{{dx}},{y^{\prime \prime }} = \frac{{{d^2}y}}{{d{x^2}}}} \right)\] will be \[y{y^{\prime \prime }} + {\left( {{y^\prime }} \right)^2} + 1 = 0\]
Hence, the option B is correct.

Note:
Student should be careful in using the formula \[\frac{d}{{dx}}(uv) = v\frac{d}{{dx}}u\] to differentiate \[\frac{{dy}}{{dx}}(y)\] by assuming \[u = \frac{{dy}}{{dx}}u = y\] because u and v are functions of \[x\]. Without employing the addition, multiplication, or quotient formulae of differentiation, the derivative or differential coefficient of a function can be directly determined by the concept of differentiation. A representation of the relationship between \[y\] and \[x\] is defined as \[\frac{{dy}}{{dx}}\].