How do you find the equation of the line tangent to the graph of y=ln⁡(5−x2) at the point where x = 2?

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abangan85s0 · Accepted Answer

We start by finding the corresponding y-coordinate that the tangent intersects the function at.
$y = \ln (5 - 2^{2})$
$y = \ln (1)$
y=0
We now find the derivative of the function.
$y = \ln (5 - x^{2})$
$e^{y} = 5 - x^{2}$
$e^{y} (\frac{d y}{d x}) = 0 - 2 x$
$\frac{d y}{d x} = - \frac{2 x}{e^{y}}$
$\frac{d y}{d x} = - \frac{2 x}{e^{\ln (5 - x^{2})}}$
$\frac{d y}{d x} = - \frac{2 x}{5 - x^{2}}$
The slope of the tangent is:
$m_{tangent} = - \frac{2 \times 2}{5 - 2^{2}}$
$m_{tangent} = - \frac{4}{1}$
$m_{tangent} = - 4$
Hence, the equation of the tangent is:
$y - y_{1} = m (x - x_{1})$
y-0=-4(x-2)
y=-4x+8

How do you find the equation of the

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