Question:
Find the second order derivatives of the function
\(x.cosx\)
Find the second order derivatives of the function
\(x.cosx\)
\(x.cosx\)
Updated On: Sep 12, 2023
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Solution and Explanation
The correct answer is \(=-(xcosx+2sinx)\)
Let \(y=x.cosx\)
Then,
\(\frac{dy}{dx}=\frac{d}{dx}(x.cosx)=cosx.\frac{d}{dx}(x)+x\frac{d}{dx}(cosx)\)
\(=cosx.1+x(-sinx)=cosx-xsinx\)
\(∴\frac{d^2y}{dx^2}=\frac{d}{dx}(cosx-xsinx)=\frac{d}{dx}(cosx)-\frac{d}{dx}(xsinx)\)
\(=-sinx-[sinx.\frac{d}{dx}(x)+x.\frac{d}{dx}(sinx)]\)
\(=-sinx-(sinx+xcosx)\)
\(=-(xcosx+2sinx)\)
Let \(y=x.cosx\)
Then,
\(\frac{dy}{dx}=\frac{d}{dx}(x.cosx)=cosx.\frac{d}{dx}(x)+x\frac{d}{dx}(cosx)\)
\(=cosx.1+x(-sinx)=cosx-xsinx\)
\(∴\frac{d^2y}{dx^2}=\frac{d}{dx}(cosx-xsinx)=\frac{d}{dx}(cosx)-\frac{d}{dx}(xsinx)\)
\(=-sinx-[sinx.\frac{d}{dx}(x)+x.\frac{d}{dx}(sinx)]\)
\(=-sinx-(sinx+xcosx)\)
\(=-(xcosx+2sinx)\)
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Second-Order Derivative
The Second-Order Derivative is the derivative of the first-order derivative of the stated (given) function. For instance, acceleration is the second-order derivative of the distance covered with regard to time and tells us the rate of change of velocity.
As well as the first-order derivative tells us about the slope of the tangent line to the graph of the given function, the second-order derivative explains the shape of the graph and its concavity.
The second-order derivative is shown using \(f’’(x)\text{ or }\frac{d^2y}{dx^2}\).