What is the tangent of f(x)=sqrt(x-1),(10,3) ?

The tangent of f(x)=sqrt(x-1),(10,3) is y= 1/6 x+4/3

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tangent of f(x)=sqrt(x-1),(10,3)

tangent of

f (x) = x - 1, (10, 3)

y = \frac{1}{6} x + \frac{4}{3}

Solution steps

Find the slope of

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

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}{6}

Find the line with slope m=

\frac{1}{6}

and passing through

(10, 3) : y = \frac{1}{6} x + \frac{4}{3}

y = \frac{1}{6} x + \frac{4}{3}

\int_{0}^{3} 4 x + 5 d x

x = - 3, x = - 1, y = x^{3} + 8, y = 0

\frac{\partial}{\partial x} (y^{2} e^{- x})

\frac{\partial}{\partial x} ((A x^{c} + B y^{c})^{\frac{1}{c}})

\frac{\partial}{\partial x} (\frac{18}{x y})

What is the tangent of f(x)=sqrt(x-1),(10,3) ?
The tangent of f(x)=sqrt(x-1),(10,3) is y= 1/6 x+4/3