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Question: If \[\sin (x+y)+\cos (x+y)=\log (x+y)\] , then \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\]is equal to A. \[...

If sin⁡(x+y)+cos⁡(x+y)=log⁡(x+y)\sin (x+y)+\cos (x+y)=\log (x+y) , then d2ydx2\dfrac{{{d}^{2}}y}{d{{x}^{2}}}is equal to
A. −yx-\dfrac{y}{x}
B. 00
C. −1-1
D. 11

Explanation

Solution

In this question firstly differentiate the given function with respect to the variable xx. If we differentiate with respect to xxthen yywill also act as a variable. After differentiation calculate the value of dydx\dfrac{dy}{dx} now again differentiate dydx\dfrac{dy}{dx}in order to get d2ydx2\dfrac{{{d}^{2}}y}{d{{x}^{2}}} . By following these steps you can get your answer.

Complete step by step answer:
The given question is of double differentiation i.e. we have to find the differentiation of differentiation of function that’s why it is known as double differentiation.
Let us understand the concept of differentiation and double differentiation in more detail. Differentiation in mathematics means the rate of change of the given function with respect to some variable. And by the definition of derivative of a function ff we can say that,
f′(x)=lim⁡h→0 f(x+h)−f(x)hf'(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}
Where, hh is the infinitesimal change in the function ff.
And the derivative of a function is denoted by the f′f' or dydx\dfrac{dy}{dx}.
Now if we talk about the second derivative then it is defined by the limit definition of the derivative of the first derivative i.e.
f′′(x)=lim⁡h→0 f′(x+h)−f′(x)hf''(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f'(x+h)-f'(x)}{h}
The second derivative of a function is denoted by the f′′f'' or d2ydx2\dfrac{{{d}^{2}}y}{d{{x}^{2}}}.
When we calculate the second derivative of the function ff then it will give the answer as the instantaneous rate of change of the first derivative. In other words we can say that the second derivative measures the rate at which the first derivative changes.
In the given question the function is sin⁡(x+y)+cos⁡(x+y)=log⁡(x+y)\sin (x+y)+\cos (x+y)=\log (x+y) we have to calculate the second derivative with respect to xx .
So, differentiate the function with respect to xx, we get
ddx(sin⁡(x+y)+cos⁡(x+y))=ddxlog⁡(x+y)\dfrac{d}{dx}(\sin (x+y)+\cos (x+y))=\dfrac{d}{dx}\log (x+y)
Using the chain rule,
⇒cos⁡(x+y)(1+dydx)−sin⁡(x+y)(1+dydx)=(1x+y)(1+dydx)\Rightarrow \cos (x+y)(1+\dfrac{dy}{dx})-\sin (x+y)(1+\dfrac{dy}{dx})=(\dfrac{1}{x+y})(1+\dfrac{dy}{dx})
Simplifying the above expression, we get
⇒(1+dydx)[cos⁡(x+y)−sin⁡(x+y)−1x+y]=0\Rightarrow (1+\dfrac{dy}{dx})\left[ \cos (x+y)-\sin (x+y)-\dfrac{1}{x+y} \right]=0
By solving the expression, we get
⇒(1+dydx)=0\Rightarrow (1+\dfrac{dy}{dx})=0
⇒dydx=−1\Rightarrow \dfrac{dy}{dx}=-1
Now we have the first derivative of the given function. But in the given question we have to calculate the second derivative. So for calculating the second derivative we have to again differentiate the first derivative. By differentiating we will get,
⇒ddxdydx=ddx(−1)\Rightarrow \dfrac{d}{dx}\dfrac{dy}{dx}=\dfrac{d}{dx}(-1)
We know that the differentiation of any constant value is zero. By using this property we get
⇒d2ydx2=0\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=0

So, the correct answer is “Option B”.

Note: We will get real life examples of derivatives from distance, speed and acceleration. The first derivative of distance with respect to time will give us Speed, while the second derivative of distance with respect to time will give us Acceleration. And the third derivative of the distance will give us Jerk. The fourth is Snap or Jounce.