Question

If $y=A\cos nx+b\sin nx$ , then the value of $\dfrac{{{d}^{2}}y}{dx{}^{2}}$is equal to ?
(a) ${{n}^{2}}y$
(b) -y
(c) -${{n}^{2}}y$
(d) none of these

Answer 1

Hint: We have given the question which is based on differentiation. The concept of differentiation deals with the idea of finding the derivative of a function. It is the process of determining the rate of change in functions on the basis of its variables. In this question, we have to differentiate the given equation two times. After differentiation, we get our desirable answer.

Formula Used: $\dfrac{d(\sin x)}{dx}=\cos x$ and $\dfrac{d(\cos x)}{dx}=-\sin x$

Complete step by step Solution:
We have given the equation $y=A\cos nx+b\sin nx$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .(1)
 We have to find out the value of $\dfrac{{{d}^{2}}y}{dx{}^{2}}$
This means we have to solve the double differentiation.
Double differentiation is the differentiation of the first derivative.
First we differentiate equation (1) w.r.t x, we get
    $\dfrac{dy}{dx}=\dfrac{d}{dx}(A\cos nx+B\sin nx)$
By expanding the above equation, we get
    $\dfrac{dy}{dx}=\dfrac{d}{dx}(A\cos nx)+\dfrac{d}{dx}(B\sin nx)$
   $\dfrac{dy}{dx}=-An\sin nx+Bn\cos nx$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
Now we differentiate equation (2) w.r.t x, we get
  $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{d{{x}^{{}}}}(-An\sin nx+Bn\cos nx)$
By expanding, we get
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}(-An\sin nx)+\dfrac{d}{dx}(Bn\cos nx)$
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-A{{n}^{2}}\cos nx-B{{n}^{2}}\sin nx$
          = $-{{n}^{2}}(A\cos nx+B\sin nx)$
As the value of y is given in the question, by substituting it in the above equation, we get
          =$-{{n}^{2}}y$
Hence, differentiation of $y=A\cos nx+b\sin nx$ is $-{{n}^{2}}y$

Hence, the correct option is C.

Note: Students make mistakes in doing differentiation. It should take care in differentiation that while differentiating the terms, we should subtract 1 from the given power. It may be noted here that differentiation is expressed in terms of parameters solely without directly involving the most variables x and y. By learning the differentiation of trigonometric terms and doing a lot of practice differentiation, students will be able to solve the question accurately.