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}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : 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}

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