Question

If $y\sec x + \tan x + {x^2}y = 0,$ then $\dfrac{{dy}}{{dx}} = $
$A)\dfrac{{(2xy + {{\sec }^2}x + y\sec x\tan x)}}{{({x^2} + \sec x)}}$
$B)\dfrac{{(2xy + {{\sec }^2}x\sec x\tan x)}}{{({x^2} + \sec x)}}$
$C)\dfrac{{ - (2xy + {{\sec }^2}x + y\sec x\tan x)}}{{({x^2} + \sec x)}}$
$D)$None of these

Answer 1

Hint: First, we need to analyze the given information which is in the trigonometric form.
The concept of trigonometric Identities are useful whenever trigonometric functions are involved in an expression or an equation and these identities are useful whenever expressions involving trigonometric functions need to be simplified.
 From the given, we asked to calculate the value of differentiation function $\dfrac{{dy}}{{dx}}$ when $y\sec x + \tan x + {x^2}y = 0,$ and we will differentiate each and every trigonometric value and variables to make the derivative part easier.
In differentiation, the derivative of $x$ raised to the power is denoted by $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$ .
Formula used:
$\dfrac{{d(\tan x)}}{{dx}} = {\sec ^2}x$
$\dfrac{{d(\sec x)}}{{dx}} =\sec x \tan x$

Complete step by step answer:
Since from the given that we have $y\sec x + \tan x + {x^2}y = 0,$we will derive this function step by step.
Let us take $y\sec x$ first, and try to derivative this part with respect to the variable $x$
Thus we get $\dfrac{{d(y\sec x)}}{{dx}} = \dfrac{{dy}}{{dx}}(\sec x) + y\sec x\tan x$ we will be done this process by using the $uv$ differentiation rule, which is if the terms are in multiplication while derivation we may apply this rule, that is derivative of $uv = {u^1}v + u{v^1}$and assume $u = y,v = \sec x$ then we have $\dfrac{{d(\sec x)}}{{dx}} =\sec x \tan x,\dfrac{{d(y)}}{{dx}} = \dfrac{{dy}}{{dx}}$ (a derivative of x terms only gets change)
Hence, we have $\dfrac{{d(y\sec x)}}{{dx}} = \dfrac{{dy}}{{dx}}(\sec x) + y\sec x\tan x$
Now take the second term $\tan x$and its derivation given as $\dfrac{{d(\tan x)}}{{dx}} = {\sec ^2}x$
Finally, take the last term ${x^2}y$and apply the same $uv = {u^1}v + u{v^1}$concept, thus we get $\dfrac{{d({x^2}y)}}{{dx}} = 2xy + {x^2}(\dfrac{{dy}}{{dx}})$
Thus, derivatives of $y\sec x + \tan x + {x^2}y = 0,$with respect to x.
Thus, we get $\dfrac{{dy}}{{dx}}(\sec x) + y\sec x\tan x + {\sec ^2}x + 2xy + {x^2}(\dfrac{{dy}}{{dx}}) = 0$
Now in the left side place, derivative function and all the terms are on the right sides, thus we have $\dfrac{{dy}}{{dx}}(\sec x) + {x^2}(\dfrac{{dy}}{{dx}}) = - [y\sec x\tan x + {\sec ^2}x + 2xy]$
Taking the common terms, we get $\dfrac{{dy}}{{dx}}(\sec x + {x^2}) = - [y\sec x\tan x + {\sec ^2}x + 2xy]$
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - [y\sec x\tan x + {{\sec }^2}x + 2xy]}}{{(\sec x + {x^2})}}$

So, the correct answer is “Option C”.

Note: Differentiation and integration are inverse processes like a derivative of \[\dfrac{{d({x^2})}}{{dx}} = 2x\] and the integration is $\int {2xdx = \dfrac{{2{x^2}}}{2}} = {x^2}$.
In total there are six trigonometric values which are sine, cos, tan, sec, cosec, cot while all the values have been relation like $\dfrac{{\sin }}{{\cos }} = \tan $ and $\tan = \dfrac{1}{{\cot }}$