What is the tangent of f(x)=ln(sqrt(x)),\at x=-1/2 ?

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

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tangent of f(x)=ln(sqrt(x)),\at x=-1/2

tangent of

f (x) = ln (x), at x = - \frac{1}{2}

y = - x + ln (i) - \frac{1}{2} ln (2) - \frac{1}{2}

Solution steps

Find the tangent point:

(- \frac{1}{2}, ln (i) - \frac{1}{2} ln (2))

Find the slope of

f (x) = ln (x) : \frac{df ( x )}{d x} = \frac{1}{2 x}

Compute the slope of

f (x) = ln (x)

x = - \frac{1}{2} : m = - 1

Find the line with slope m=

- 1

and passing through

(- \frac{1}{2}, ln (i) - \frac{1}{2} ln (2)) : y = - x + ln (i) - \frac{1}{2} ln (2) - \frac{1}{2}

y = - x + ln (i) - \frac{1}{2} ln (2) - \frac{1}{2}

\frac{d x}{d t} = x^{2} + \frac{1}{25}

y = 6 x + 4 x - x^{2}, y = 2 x - 2

f (x) = x^{6} cos (x)

\frac{d}{d x} (x^{2} 10 x - 7)

f (x) = (x - 1) e^{x}

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