The curve f(x) = e^x sin x is defined in the | vedclass

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(B) Given the function $f(x) = e^x \sin x$. The slope of the tangent at any point $x$ is given by $f'(x)$. $f'(x) = \frac{d}{dx}(e^x \sin x) = e^x \si

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$\frac{\pi}{2}$

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(B) Given the function $f(x) = e^x \sin x$. The slope of the tangent at any point $x$ is given by $f'(x)$. $f'(x) = \frac{d}{dx}(e^x \sin x) = e^x \sin x + e^x \cos x = e^x(\sin x + \cos x)$. To find the maximum slope,we need to find the critical points of $f'(x)$ by setting its derivative $f''(x) = 0$. $f''(x) = \frac{d}{dx}(e^x \sin x + e^x \cos x) = (e^x \sin x + e^x \cos x) + (e^x \cos x - e^x \sin x) = 2e^x \cos x$. Setting $f''(x) = 0$,we get $2e^x \cos x = 0$. Since $e^x 
eq 0$ for all $x$,we must have $\cos x = 0$. In the interval $[0, 2\pi]$,$\cos x = 0$ at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$. Checking the second derivative test or evaluating the slope,the slope $f'(x) = e^x(\sin x + \cos x)$ is maximum at $x = \frac{\pi}{2}$.

The curve $f(x) = e^x \sin x$ is defined in the interval $[0, 2\pi]$. The value of $x$ for which the slope of the tangent drawn to the curve at $x$ is maximum,is

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