The equation of the normal to the curve y = l | vedclass

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Question

(A) Given the curve $y = \log_e x$. First,find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent at point $P(1, 0)$. $\frac{dy}{dx}

Vedclass · Accepted Answer

$x + y = 1$

Vedclass · Answer

(A) Given the curve $y = \log_e x$. First,find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent at point $P(1, 0)$. $\frac{dy}{dx} = \frac{1}{x}$. At point $P(1, 0)$,the slope of the tangent $m_t = \left. \frac{dy}{dx} \right|_{(1, 0)} = \frac{1}{1} = 1$. The slope of the normal $m_n$ is given by $-\frac{1}{m_t} = -\frac{1}{1} = -1$. The equation of the normal line passing through $(x_1, y_1) = (1, 0)$ with slope $m_n = -1$ is: $y - y_1 = m_n(x - x_1)$ $y - 0 = -1(x - 1)$ $y = -x + 1$ $x + y = 1$.

The equation of the normal to the curve $y = \log_e x$ at the point $P(1, 0)$ is $ . . . . . . $

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