Question:
Find the second order derivatives of the function
\(sin(logx)\)
Find the second order derivatives of the function
\(sin(logx)\)
\(sin(logx)\)
Updated On: Sep 12, 2023
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Solution and Explanation
The correct answer is \(=\frac{-[sin(logx)+cos(logx)]}{x^2}\)
Let \(y=sin(logx)\)
Then,
\(\frac{dy}{dx}=\frac{d}{dx}(sin(logx))=cos(logx).\frac{d}{dx}(logx)=\frac{cos(logx)}{x}\)
\(∴\frac{d^2y}{dx^2}=\frac{d}{dx}[\frac{cos(logx)}{x}]\)
\(=\frac{x.\frac{d}{dx}[cos(logx)]-cos(logx).\frac{d}{dx}(x)}{x^2}\)
\(=\frac{x.[-sin(logx).\frac{d}{dx}(logx)]-cos(logx).1}{x^2}\)
\(=\frac{-xsin(logx).\frac{1}{x}-cos(logx)}{x^2}\)
\(=\frac{-[sin(logx)+cos(logx)]}{x^2}\)
Let \(y=sin(logx)\)
Then,
\(\frac{dy}{dx}=\frac{d}{dx}(sin(logx))=cos(logx).\frac{d}{dx}(logx)=\frac{cos(logx)}{x}\)
\(∴\frac{d^2y}{dx^2}=\frac{d}{dx}[\frac{cos(logx)}{x}]\)
\(=\frac{x.\frac{d}{dx}[cos(logx)]-cos(logx).\frac{d}{dx}(x)}{x^2}\)
\(=\frac{x.[-sin(logx).\frac{d}{dx}(logx)]-cos(logx).1}{x^2}\)
\(=\frac{-xsin(logx).\frac{1}{x}-cos(logx)}{x^2}\)
\(=\frac{-[sin(logx)+cos(logx)]}{x^2}\)
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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 \(f’’(x)\text{ or }\frac{d^2y}{dx^2}\).