What is the tangent of f(x)=sqrt(x),(64,8) ?

The tangent of f(x)=sqrt(x),(64,8) is y= 1/16 x+4

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tangent of f(x)=sqrt(x),(64,8)

tangent of

f (x) = x, (64, 8)

y = \frac{1}{16} x + 4

Solution steps

Find the slope of

f (x) = x : \frac{df ( x )}{d x} = \frac{1}{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}{16}

Find the line with slope m=

\frac{1}{16}

and passing through

(64, 8) : y = \frac{1}{16} x + 4

y = \frac{1}{16} x + 4

\frac{\partial}{\partial x} (cos (x - y))

\frac{d}{d x} (21 x e^{- 2 x})

f (x) = 2 x + 4

\int 2 x (x^{2} + 9)^{7} d x

\int \frac{5}{9 x ^{2} - 16} d x

What is the tangent of f(x)=sqrt(x),(64,8) ?
The tangent of f(x)=sqrt(x),(64,8) is y= 1/16 x+4