Question

If $ y = x + \tan x $ , prove that $ {\cos ^2}x\;{y_2} - 2y + 2x = 0 $

Answer 1

Hint: This question is based on differentiation of the basic trigonometric functions. Also, the differential of term raised to a certain power can be expressed as $ \dfrac{d}{{dx}}{\left( x \right)^n} = n{x^{n - 1}} $ .
When we differentiate a term raised to certain power we bring its power down and in front of the term and also will reduce its power by one. To prove the result, we will differentiate the expression given in the question with respect to x. We will again differentiate the expression with respect to $ x $ . The differential of $ \sec x $ is written as $ \sec x\tan x $ . Then we will substitute \[\dfrac{1}{{{{\cos }^2}x}}\] for $ {\sec ^2}x $ as $ \cos x $ is the inverse of $ secx $ . Then we will rearrange the expression to get the desired result.

Complete step-by-step answer:
Given:
The given expression is $ y = x + \tan x $ .
The expression that we need to get is in the double differential form. Hence, we will differentiate the above expression with respect to $ x $ .
 $ \dfrac{{dy}}{{dx}} = 1 + {\sec ^2}x $
We will again differentiate the above expression with respect to $ x $ . We get,
 $ \begin{array}{l}
\dfrac{{{d^2}y}}{{d{x^2}}} = 2\sec x \times \left( {\sec x\tan x} \right)\\
\dfrac{{{d^2}y}}{{d{x^2}}} = 2{\sec ^2}x\tan x
\end{array} $

We will write $ \dfrac{1}{{{{\cos }^2}x}} $ for $ {\sec ^2}x $ in the above expression as $ \cos x $ is the inverse of $ secx $ .
 $ \dfrac{{{d^2}y}}{{d{x^2}}} = 2\dfrac{1}{{{{\cos }^2}x}} \times \tan x $
 We will rearrange the above expression as:
 $ {\cos ^2}x\dfrac{{{d^2}y}}{{d{x^2}}} - 2\tan x = 0 $
Now we will add and subtract $ 2x $ in the left hand side of the above expression.
 $ \begin{array}{l}
{\cos ^2}x\dfrac{{{d^2}y}}{{d{x^2}}} - 2x - 2\tan x + 2x = 0\\
{\cos ^2}x\dfrac{{{d^2}y}}{{d{x^2}}} + 2x - 2\left( {x + \tan x} \right) = 0
\end{array} $
From the question we know that, $ y = x + \tan x $ . So, we will substitute $ y $ for $ x + \tan x $ in the above expression.
 $ {\cos ^2}x\dfrac{{{d^2}y}}{{d{x^2}}} + 2x - 2y = 0 $
Hence, it is proved.

Note: Before solving this question, we need to revise all the formulas of trigonometric function. Since, this question deals with the differentiation of trigonometric functions. Also the differentiation of constant terms always equals to zero. Second differential is purely based on the first differential hence the first differential needs to be exactly calculated with proper sign conventions.