Question

Differential equation of \[y = \sec \left( {{{\tan }^{ - 1}}x} \right)\] is
A. \[\left( {1 + {x^2}} \right)\frac{{dy}}{{dx}} = y + x\]
В. \[\left( {1 + {x^2}} \right)\frac{{dy}}{{dx}} = y - x\]
C. \[\left( {1 + {x^2}} \right)\frac{{dy}}{{dx}} = xy\]
D. \[\left( {1 + {x^2}} \right)\frac{{dy}}{{dx}} = \frac{x}{y}\]

Answer 1

Hint:
Consider '\[y\]' as composite function of \[f(g(x))\] where \[f(x)\]is \[\sec x\] and \[g(x)\] is \[{\tan ^{ - 1}}x\] and then use the identity
\[\frac{d}{{dx}}(f(g(x))) = {f^\prime }(g(x)) \times {g^\prime }(x)\]
Where \[{f^\prime }(g(x))\] is differentiation of \[{\rm{f}}({\rm{x}})\] keeping \[g(x)\] as it is and \[{g^\prime }(x)\] means differentiating\[g(x)\] irrespective of \[f(x)\].
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:
In this question, we have been given the equation
\[y = \sec \left( {{{\tan }^{ - 1}}x} \right)\]
Now, for the above equation, find \[\frac{{dy}}{{dx}}\].
That is differentiating the above equation of ‘y’ with respect to ‘x’.
We have to use the identity: \[\frac{d}{{dx}}(f(g(x))) = {f^\prime }(g(x)) \times {g^\prime }(x)\]
Here, \[{f^\prime }(g(x))\] means differentiating \[f(x)\] keeping \[g(x)\] constant and here \[{g^\prime }(x)\] means differentiating \[{g^\prime }(x)\] independently irrespective of \[f(x)\].
So we have to use the formula that are listed below,
\[\frac{d}{{dx}}(\sec x) = \sec x\tan x,\frac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right) = \frac{1}{{1 + {x^2}}}\]
On differentiating the given equation using the above formula, we get
\[\frac{{dy}}{{dx}} = \sec \left( {{{\tan }^{ - 1}}x} \right).\tan \left( {{{\tan }^{ - 1}}x} \right) \cdot \frac{1}{{1 + {x^2}}}\]---- (1)
Now, the identity that is listed previously can be used here:
The identities are as below:
\[\tan \left( {{{\tan }^{ - 1}}x} \right) = x\]
\[\sec \left( {{{\tan }^{ - 1}}x} \right) = \sqrt {1 + {x^2}} \]
By using the above identities, the equation (1) can be written as
\[\frac{{dy}}{{dx}} = \sqrt {1 + {x^2}} \cdot x \cdot \frac{1}{{1 + {x^2}}}\]
Now, by rationalizing the above equation, we get
\[ \Rightarrow \frac{{dy}}{{dx}} = \frac{x}{{\sqrt {1 + {x^2}} }}\]
Therefore, the required differential equation of \[y = \sec \left( {{{\tan }^{ - 1}}x} \right)\] is \[\left( {1 + {x^2}} \right)\frac{{dy}}{{dx}} = xy\]
Hence the option C is correct.
Note:
There is alternative way of solving the problem is by converting \[y = \sec \left( {{{\tan }^{ - 1}}x} \right)\]as \[y = \sqrt {1 + {x^2}} \] and using \[f(x) = \sqrt x \]and\[g(x) = \left( {1 + {x^2}} \right)\]. Student mostly makes mistakes in determining the inverse of trigonometry function. According to the rule of trigonometric inverse functions, the first thing to keep in mind is that \[\left( {{{\tan }^{ - 1}}x} \right) = x\], but \[{\tan ^{ - 1}}(\tan x) = x\] if and only if it is mentioned that\[x \in \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)\]. Another thing to remember is that when reporting the answers, report the exact matched option, as the options provided are very similar and misleading.