What is the tangent of y=ln(sqrt(3-2x)),\at x=-3 ?

The tangent of y=ln(sqrt(3-2x)),\at x=-3 is y=-1/9 x+ln(3)-1/3

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tangent of y=ln(sqrt(3-2x)),\at x=-3

tangent of

y = ln (3 - 2 x), at x = - 3

y = - \frac{1}{9} x + ln (3) - \frac{1}{3}

Solution steps

Find the tangent point:

(- 3, ln (3))

Find the slope of

y = ln (3 - 2 x) : \frac{d y}{d x} = - \frac{1}{3 - 2 x}

Compute the slope of

y = ln (3 - 2 x)

x = - 3 : m = - \frac{1}{9}

Find the line with slope m=

- \frac{1}{9}

and passing through

(- 3, ln (3)) : y = - \frac{1}{9} x + ln (3) - \frac{1}{3}

y = - \frac{1}{9} x + ln (3) - \frac{1}{3}

\int ln (x + 3) d x

\frac{d}{d x} ((3 x - 2)^{3})

y = 9 x - x^{2}, y = 2 x

\int x^{7} e^{- x^{8}} d x

y^{^{'}} + \frac{1}{x} y = x^{2} y^{1}

What is the tangent of y=ln(sqrt(3-2x)),\at x=-3 ?
The tangent of y=ln(sqrt(3-2x)),\at x=-3 is y=-1/9 x+ln(3)-1/3