Question:
Find the second order derivatives of the function
\(x^3logx\)
Find the second order derivatives of the function
\(x^3logx\)
\(x^3logx\)
Updated On: Sep 12, 2023
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Solution and Explanation
The correct answer is \(=x(5+6\,logx)\)
Let \(y=x^3logx\)
Then,
\(\frac{dy}{dx}=\frac{d}{dx}(x^3logx)=logx.\frac{d}{dx}(x^3)+x^3.\frac{d}{dx}(logx)\)
\(=logx.3x^2+x^3.\frac{1}{x}=logx.3x^2+x^2\)
\(=x^2(1+3logx)\)
\(∴\frac{d^2y}{dx^2}=\frac{d}{dx}[x^2(1+3logx)]\)
\(=(1+3logx).\frac{d}{dx}(x^2)+x^2\frac{d}{dx}(1+3logx)\)
\(=(1+3logx).2x+x^2.\frac{3}{x}\)
\(=2x+6x\,logx+3x\)
\(=5x+6x\,logx\)
\(=x(5+6\,logx)\)
Let \(y=x^3logx\)
Then,
\(\frac{dy}{dx}=\frac{d}{dx}(x^3logx)=logx.\frac{d}{dx}(x^3)+x^3.\frac{d}{dx}(logx)\)
\(=logx.3x^2+x^3.\frac{1}{x}=logx.3x^2+x^2\)
\(=x^2(1+3logx)\)
\(∴\frac{d^2y}{dx^2}=\frac{d}{dx}[x^2(1+3logx)]\)
\(=(1+3logx).\frac{d}{dx}(x^2)+x^2\frac{d}{dx}(1+3logx)\)
\(=(1+3logx).2x+x^2.\frac{3}{x}\)
\(=2x+6x\,logx+3x\)
\(=5x+6x\,logx\)
\(=x(5+6\,logx)\)
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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 \(f’’(x)\text{ or }\frac{d^2y}{dx^2}\).