Question

If $ y = \tan x, $ then $ \dfrac{{{d^2}y}}{{d{x^2}}} = $
A. $ x\dfrac{{dy}}{{dx}} $
B. $ 2x\dfrac{{dy}}{{dx}} $
C. $ 2y\dfrac{{dy}}{{dx}} $
D. $ y\dfrac{{dy}}{{dx}} $

Answer 1

Hint: Here, we will follow the step by step approach, first find the first order derivative, then substitute using the basic trigonometric function and convert the equation in the form of the given expression and then find the second order derivative.

Complete step-by-step answer:
Take the given equation –
 $ y = \tan x $ .... (A)
Take derivative with respect to “x” on both the sides of the equations
 $ \Rightarrow \dfrac{d}{{dx}}(y) = \dfrac{d}{{dx}}(\tan x) $
Place the formula using derivative of tangent on the right hand side of the equation –
 $ \Rightarrow \dfrac{{dy}}{{dx}} = {\sec ^2}x $
Using the trigonometric identity which states that $ 1 + {\tan ^2}x = {\sec ^2}x $ , place it in the above equation.
 $ \Rightarrow \dfrac{{dy}}{{dx}} = 1 + {\tan ^2}x $
By using the equation (A), we can rewrite the above equation as –
 $ \Rightarrow \dfrac{{dy}}{{dx}} = 1 + {y^2} $
Take again derivative on both the sides of the above equation
 $ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{d}{{dx}}\left( {1 + {y^2}} \right) $
The above equation gives derivative of the second order
 $ \Rightarrow \left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right) = \dfrac{d}{{dx}}\left( {1 + {y^2}} \right) $
Now, take the derivative on the right hand side of the equation and apply by using the distributive property among the additions.
 $ \Rightarrow \left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right) = \dfrac{d}{{dx}}(1) + \dfrac{d}{{dx}}({y^2}) $
Derivative of the constant term is always zero.
 $ \Rightarrow \left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right) = 0 + \dfrac{d}{{dx}}({y^2}) $
Now apply the formula for the derivative – $ \dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}} $ and using the chain rule in the derivative.
 $ \Rightarrow \left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right) = 2y\dfrac{d}{{dx}}(y) $
The above equation can be re-written as –
 $ \Rightarrow \left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right) = 2y\dfrac{{dy}}{{dx}} $
Hence, from the given multiple choices- the option C is the correct answer.
So, the correct answer is “Option C”.

Note: Always remember the basic trigonometric and the derivative formulas for the accurate and efficient solution. Remember the relation between the trigonometric functions to substitute the values in and during the solution. Also, go through the chain rule of the derivatives and the first, second and third order of the derivative.