If y = cos x then \(\rm \frac{d^2y}{dx^2} + y =\)1. y2. -y3. 04. 1

1.	If y = cos x then \(\rm \frac{d^2y}{dx^2} + y =\)1. y2. -y3. 04. 1
Answer» Correct Answer - Option 3 : 0 Concept: Suppose that we have two functions f(x) and g(x) and they are both differentiable. Chain Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right)} \right] = {\rm{\;f'}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right){\rm{g'}}\left( {\rm{x}} \right)\) Product Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right)} \right] = {\rm{\;f'}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right) + {\rm{f}}\left( {\rm{x}} \right){\rm{\;g'}}\left( {\rm{x}} \right)\) Formulas: \(\rm\frac{d\sin x}{dx} = \cos x\) \(\rm\frac{d\cos x}{dx} =- \sin x\) Calculation: We have to find the value of \(\rm \frac{d^2y}{dx^2}\) Given: y = cos x \(\rm \frac{d^2y}{dx^2} =\rm \frac{d^2\cos x}{dx^2} = \frac{d}{dx} \left(\frac{d\cos x}{dx} \right )\) \(= \rm\frac{d(-\sin x)}{dx} =- \cos x\) \(\rm \frac{d^2y}{dx^2}\) = -y ∴ \(\rm \frac{d^2y}{dx^2} + y = 0\)

Answer» Correct Answer - Option 3 : 0

Concept:

Suppose that we have two functions f(x) and g(x) and they are both differentiable.

Chain Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right)} \right] = {\rm{\;f'}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right){\rm{g'}}\left( {\rm{x}} \right)\)
Product Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right)} \right] = {\rm{\;f'}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right) + {\rm{f}}\left( {\rm{x}} \right){\rm{\;g'}}\left( {\rm{x}} \right)\)

Formulas:

\(\rm\frac{d\sin x}{dx} = \cos x\)

\(\rm\frac{d\cos x}{dx} =- \sin x\)

Calculation:

We have to find the value of \(\rm \frac{d^2y}{dx^2}\)

Given: y = cos x

\(\rm \frac{d^2y}{dx^2} =\rm \frac{d^2\cos x}{dx^2} = \frac{d}{dx} \left(\frac{d\cos x}{dx} \right )\)

\(= \rm\frac{d(-\sin x)}{dx} =- \cos x\)

\(\rm \frac{d^2y}{dx^2}\) = -y

∴ \(\rm \frac{d^2y}{dx^2} + y = 0\)

Discussion