Find the second order derivatives of the function
\(e^xsin5x\)
\(e^xsin5x\)
Solution and Explanation
Let \(y=e^xsin5x\)
Then,
\(\frac{dy}{dx}=\frac{d}{dx}(e^xsin5x)=sin5x.\frac{d}{dx}(e^x)+e^x\frac{d}{dx}(sin5x)\)
\(=sin5x.e^x+e^x.cos5x\frac{d}{dx}(5x)=e^xsin5x+e^xcos5x.5\)
\(=e^x(sin5x+5cos5x)\)
\(∴\frac{d^2y}{dx^2}=\frac{d}{dx}[e^x(sin5x+5cos5x)]\)
\(=(sin5x+5cos5x).\frac{d}{dx}(e^x)+e^x.\frac{d}{dx}(sin5x+5cos5x)\)
\(=(sin5x+5cos5x)e^x+e^x[cos5x.\frac{d}{dx}(5x)+5(-sin5x).\frac{d}{dx}(5x)]\)
\(=e^x(sin5x+5cos5x)+e^x(5cos5x-25sin5x)\)
Then, \(e^x(10cos5x-24sin5x)=2e^x(5cos5x-12sin5x)\)
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Concepts Used:
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 \(f’’(x)\text{ or }\frac{d^2y}{dx^2}\).