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Q.
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
We have,
$f(x)=e^{x} \cdot \sin x$
difference w.r.t ' $x^{'}$
$f^{'}(x)=e^{x} \cos x+e^{x} \sin x$
For $f^{'}(x)$ is maximum we find $f^{''}(x)$
$f^{''}(x)=-e^{x} \sin x+e^{x} \cos x+e^{x} \cos x +e^{x} \sin x=2 e^{x} \cos x$
$\therefore f^{'}(x)$ is maximum at $f^{''}(x)=0$
Now, $f^{''}(x)=0$
$\therefore 2 e^{x} \cos x=0$
$\Rightarrow x=\frac{\pi}{2}$
So, slope of tangent drawn to curve is maximum
at $x=\frac{\pi}{2}$.