Find \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\]
\[y=x\cos x\]
Answer
676.2k+ views
Hint:If u and v are two differentiable functions of x then \[\dfrac{d}{dx}\left( uv \right)=u\times \dfrac{dv}{dx}+v\times \dfrac{du}{dx}\]and this formula is called product rule. In this problem u and v are two differentiable functions of x so apply the product rule. Here they asked us to find the \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\]so we have to apply the product rule two times then we will get the required answer.
Complete step-by-step answer:
Given that \[y=x\cos x\]
We have to find \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\]
We know that the formula for \[\dfrac{d}{dx}\left( uv \right)\]is given by \[\dfrac{d}{dx}\left( uv \right)=v\times \dfrac{du}{dx}+u\times \dfrac{dv}{dx}\]
Now apply the above formula we will get,
\[\dfrac{dy}{dx}=x\left( -\sin x \right)+\cos x(1)\]
\[\dfrac{dy}{dx}=-x\sin x+\cos x\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
Now again apply derivative to it then we will get the second derivative or double derivative.
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)\]
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( -x\sin x+\cos x \right)\]. . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( -x\sin x \right)+\dfrac{d}{dx}\left( \cos x \right)\]. . . . . . . . . . . . . . . . . . . . . . (3)
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-x(\cos x)+\sin x(-1)-\sin x\]. . . . . . . . . . . . . . . . . . . . .(4)
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-x\cos x-2\sin x\]
Note:In the above problem in the equation (3) we have used the formula that is for \[\dfrac{d}{dx}\left( u+v \right)=\dfrac{du}{dx}+\dfrac{dv}{dx}\]and this formula is also called as sum rule. The derivative of \[\sin x\]is \[\cos x\]and the derivative of \[\cos x\]is \[-\sin x\]. So we should be keen on basic trigonometric formulas and their derivative formulas for doing this problem. Carefully do the basic mathematical operations like addition, subtraction then we will get the required answer.
Complete step-by-step answer:
Given that \[y=x\cos x\]
We have to find \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\]
We know that the formula for \[\dfrac{d}{dx}\left( uv \right)\]is given by \[\dfrac{d}{dx}\left( uv \right)=v\times \dfrac{du}{dx}+u\times \dfrac{dv}{dx}\]
Now apply the above formula we will get,
\[\dfrac{dy}{dx}=x\left( -\sin x \right)+\cos x(1)\]
\[\dfrac{dy}{dx}=-x\sin x+\cos x\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
Now again apply derivative to it then we will get the second derivative or double derivative.
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)\]
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( -x\sin x+\cos x \right)\]. . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( -x\sin x \right)+\dfrac{d}{dx}\left( \cos x \right)\]. . . . . . . . . . . . . . . . . . . . . . (3)
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-x(\cos x)+\sin x(-1)-\sin x\]. . . . . . . . . . . . . . . . . . . . .(4)
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-x\cos x-2\sin x\]
Note:In the above problem in the equation (3) we have used the formula that is for \[\dfrac{d}{dx}\left( u+v \right)=\dfrac{du}{dx}+\dfrac{dv}{dx}\]and this formula is also called as sum rule. The derivative of \[\sin x\]is \[\cos x\]and the derivative of \[\cos x\]is \[-\sin x\]. So we should be keen on basic trigonometric formulas and their derivative formulas for doing this problem. Carefully do the basic mathematical operations like addition, subtraction then we will get the required answer.
Recently Updated Pages
What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

10 examples of friction in our daily life

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

A Paragraph on Pollution in about 100-150 Words

Trending doubts
What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

Explain the energy losses in the transformer How are class 12 physics CBSE

Draw a well labelled diagram of the human brain class 12 biology CBSE

Differentiate between internal fertilization and external class 12 biology CBSE

State the principle of an ac generator and explain class 12 physics CBSE

What is the Full Form of 1.DPT 2.DDT 3.BCG

