Question

If \[y = \cot x\], show that \[\dfrac{{{d^2}y}}{{d{x^2}}} + 2y\dfrac{{dy}}{{dx}} = 0\]

Answer 1

Hint: In this question, we have to evaluate the value of the left hand side to show that left hand side equals the right hand side using the given particular function. For that first we have to find the first and second derivative of the given function. By substituting the derivative values into the required result term, we will get the solution.
First we have to differentiate the given function with respect to x to find the first derivative of the given function. And then differentiate it again to find the second derivative of the given function and putting all the values in the left hand side, we will get the required result.

Formula used: Differentiation formula:
\[\dfrac{d}{{dx}}\left( {\cot x} \right) = - \cos e{c^2}x\]
\[\dfrac{d}{{dx}}\left( {\cos ecx} \right) = - \cos ecx\cot x\]
\[\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}\]

Complete step-by-step answer:
It is given that, \[y = \cot x\].
We need to show that, \[\dfrac{{{d^2}y}}{{d{x^2}}} + 2y\dfrac{{dy}}{{dx}} = 0\].
Now, \[y = \cot x.................i)\]
Differentiating both sides with respect to x, we get,
$\Rightarrow$\[\dfrac{{dy}}{{dx}} = - \cos e{c^2}x……….........ii)\]
Differentiating both sides of ii) with respect to x, we get,
$\Rightarrow$\[\dfrac{{{d^2}y}}{{d{x^2}}} = - 2\cos ecx\left( { - \cos ecx\cot x} \right)\]
Simplifying we get,
$\Rightarrow$\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 2\cos e{c^2}x\cot x...................iii)\]
Now putting all the values in left hand side from i), ii) and iii) we get,
L.H.S.=
\[\dfrac{{{d^2}y}}{{d{x^2}}} + 2y\dfrac{{dy}}{{dx}}\]
We have find that for the given term is,
$\Rightarrow$\[\dfrac{{dy}}{{dx}} = - \cos e{c^2}x\] and \[\dfrac{{{d^2}y}}{{d{x^2}}} = 2\cos e{c^2}x\cot x\]
By substituting the values into required relation we get,
\[ \Rightarrow 2\cos e{c^2}x\cot x + 2\cot x \times \left( { - \cos e{c^2}x} \right)\]
Multiplying the term that is in a bracket we get,
\[ \Rightarrow 2\cos e{c^2}x\cot x - 2\cos e{c^2}x\cot x\]
Subtracting the terms,
\[ \Rightarrow 0\]
=R.H.S.
Therefore we get, left hand side =right hand side.
Hence, \[\dfrac{{{d^2}y}}{{d{x^2}}} + 2y\dfrac{{dy}}{{dx}} = 0\]
Hence proved.

Note: The derivative of a function of a real variable measures the sensitivity to change of the function value with respect to a change in its argument. Derivatives are a fundamental tool of calculus.
The first Derivative of a function \[y = f\left( x \right)\] can be written as \[\dfrac{{dy}}{{dx}}\] or \[f'\left( x \right)\].
The first derivative primarily tells us about the direction the function is going. That is, it tells us if the function is increasing or decreasing. The first derivative can be interpreted as an instantaneous rate of change. The first derivative can also be interpreted as the slope of the tangent line.
The second derivative is the rate of change of the rate of change of a point at a graph (the "slope of the slope" if you will). This can be used to find the acceleration of an object (velocity is given by first derivative).
The second Derivative of a function \[y = f\left( x \right)\] can be written as \[\dfrac{{{d^2}y}}{{d{x^2}}}\] or \[f''\left( x \right)\].