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