Question

Find the second derivative of the following function. \[y=\cos (2x+3)\].

Answer 1

Hint: Differentiate the given function twice in order to get the second derivative. Use chain rule of composition of two functions, i.e., \[\dfrac{dy}{dx}=\dfrac{dg(h(x))}{dh(x)}\times \dfrac{dh(x)}{dx}\] to get the first derivative. Also ese the fact that differentiation of function of the form \[y=\cos x\] is \[\dfrac{dy}{dx}=-\sin x\] and \[y=ax+b\] is \[\dfrac{dy}{dx}=a\]. Then differentiate the first derivative using the product rule of differentiation and get the second derivative.

Complete step by step answer:
To find the second derivative of the function \[y=\cos (2x+3)\], we will differentiate it twice with respect to the variable x.
To find the first derivative, we will differentiate the function y with respect to x.
We will use chain rule of differentiation of composition of two functions to find the derivative which states that if y is a composition of two functions \[g(x)\] and \[h(x)\] such that \[y=g(h(x))\], then \[\dfrac{dy}{dx}=\dfrac{dg(h(x))}{dh(x)}\times \dfrac{dh(x)}{dx}\].
Substituting \[g(x)=\cos (x)\] and \[h(x)=2x+3\] in the above equation, we get \[\dfrac{dy}{dx}=\dfrac{d\cos (2x+3)}{d(2x+3)}\times \dfrac{d}{dx}(2x+3).....\left( 1 \right)\].
We know that differentiation of any function of the form \[y=ax+b\] is such that \[\dfrac{dy}{dx}=a\].
Substituting \[a=2,b=3\] in the above equation, we get \[\dfrac{d}{dx}h(x)=\dfrac{d}{dx}(2x+3)=2.....\left( 2 \right)\].
To find the value of \[\dfrac{d\cos (2x+3)}{d(2x+3)}\], let’s assume \[t=2x+3\].
Thus, we have \[\dfrac{d\cos (2x+3)}{d(2x+3)}=\dfrac{d}{dt}\cos (t)\] .
We know that differentiation of any function of the form \[y=\cos x\] is such that \[\dfrac{dy}{dx}=-\sin x\].
So, we will get \[\dfrac{d\cos (2x+3)}{d(2x+3)}=\dfrac{d}{dt}\cos (t)=-\sin (t)=-\sin (2x+3).....\left( 3 \right)\].
Substituting equation \[(2)\] and \[(3)\] in equation \[(1)\], we get \[\dfrac{dy}{dx}=\dfrac{d}{dx}\cos (2x+3)=\dfrac{d\cos (2x+3)}{d(2x+3)}\times \dfrac{d(2x+3)}{dx}=-\sin (2x+3)\times 2=-2\sin (2x+3)\].
Thus, we have \[\dfrac{dy}{dx}=\dfrac{d\cos (2x+3)}{dx}=-2\sin (2x+3)\] as the first derivative of the given function.
Now, to find the second derivative, we will again differentiate the function \[\dfrac{dy}{dx}\] as \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)\].
Let’s assume \[f(x)=\dfrac{dy}{dx}=-2\sin (2x+3)\].
So, we have \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}f(x)\].
We can rewrite \[f(x)\] as a product of two functions \[u(x)\] and \[v(x)\] such that \[u(x)=-2\] and \[v(x)=\sin (2x+3)\].
We will use the product rule of differentiation of product of two functions which states that \[\dfrac{d}{dx}f(x)=\dfrac{d}{dx}\left( u(x)\times v(x) \right)=v(x)\times \dfrac{d}{dx}u(x)+u(x)\dfrac{d}{dx}v(x).....\left( 4 \right)\].
We will now differentiate the functions \[u(x)\] and \[v(x)\].
We know that differentiation of a constant with respect to any variable is 0.
So, we have \[\dfrac{d}{dx}u(x)=\dfrac{d}{dx}(-2)=0.....\left( 5 \right)\].
As \[v(x)=\sin (2x+3)\] is a composition of two functions \[h(x)=2x+3\] and \[w(x)=\sin (x)\] such that \[v(x)=w(h(x))\], we will differentiate it according to the chain rule of differentiation of composition of two functions.
Thus, we have \[\dfrac{d}{dx}v(x)=\dfrac{d}{dx}w(h(x))=\dfrac{dw(h(x))}{dh(x)}\times \dfrac{d}{dx}h(x)\].
Substituting \[v(x)=\sin (2x+3)\], \[h(x)=2x+3\] and \[w(x)=\sin (x)\] in the above equation, we have \[\dfrac{dv(x)}{dx}=\dfrac{d}{dx}\sin (2x+3)=\dfrac{d\sin (2x+3)}{d(2x+3)}\times \dfrac{d}{dx}(2x+3).....\left( 6 \right)\].
Using equation \[(2)\], we have \[\dfrac{d}{dx}h(x)=\dfrac{d}{dx}(2x+3)=2\].
To find the value of \[\dfrac{d\sin (2x+3)}{d(2x+3)}\], we will assume \[t=2x+3\].
So, we have \[\dfrac{d\sin (2x+3)}{d(2x+3)}=\dfrac{d\sin (t)}{dt}\].
We know that differentiation of any function of the form \[y=\sin (x)\] is such that \[\dfrac{dy}{dx}=\cos (x)\].
Thus, we have \[\dfrac{d\sin (2x+3)}{d(2x+3)}=\dfrac{d\sin (t)}{dt}=\cos (t)=\cos (2x+3).....\left( 7 \right)\].
Substituting the value of equation \[(2)\] and \[(7)\] in equation \[(6)\], we have \[\begin{align}
  & \dfrac{dv(x)}{dx}=\dfrac{d}{dx}\sin (2x+3)=\dfrac{d\sin (2x+3)}{d(2x+3)}\times \dfrac{d}{dx}(2x+3) \\
 & \dfrac{dv(x)}{dx}=\cos (2x+3)\times 2=2\cos (2x+3).....\left( 8 \right) \\
\end{align}\].
We will substitute \[u(x)=-2\], \[v(x)=\sin (2x+3)\] and equation \[(5)\] and \[(8)\] in equation \[(4)\].
So, we have \[\dfrac{d}{dx}f(x)=\sin (2x+3)\times (0)+(-2)\times (2\cos (2x+3))=-4\cos (2x+3)\].
Thus, we have \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}f(x)=-4\cos (2x+3)\].
Hence, the second derivative of \[y=\cos (2x+3)\] is \[-4\cos (2x+3)\].

Note:
We can also solve this question using the First Principle of Differentiation. It is necessary to use the chain rule of differentiation for composition of two functions. Otherwise, we won’t be able to find the second derivative of the given function.