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Q.
The equation of the tangent to the curve $y = 1 - e^{x/2}$ at the point of intersection with the y-axis is:
Application of Derivatives
Solution:
Consider the equation of the curve;
$y = 1 - e^{\frac{x}{2}}$
Differentiate both side w.r.t. 'x'
$\frac{dy}{dx} = - \frac{1}{2} e^{\frac{x}{2}}$
The point of intersection of the curve with y-axis is x = 0, y = 0
$\therefore \, \, \frac{dy}{dx} \bigg|_{(0,0)} = \frac{-1}{2}$
$\therefore $ Equation of the tangent to the curve is given by
$(y - 0) = \frac{dy}{dx} (x -0)$
$\Rightarrow \, y = \frac{-1}{2} (x) \, $
$\Rightarrow \, 2y + x = 0 $