What is the tangent of f(x)=sqrt(x+3),\at x=6 ?

The tangent of f(x)=sqrt(x+3),\at x=6 is y= 1/6 x+2

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tangent of f(x)=sqrt(x+3),\at x=6

tangent of

f (x) = x + 3, at x = 6

y = \frac{1}{6} x + 2

Solution steps

Find the tangent point:

(6, 3)

Find the slope of

f (x) = x + 3 : \frac{df ( x )}{d x} = \frac{1}{2 x + 3}

Compute the slope of

f (x) = x + 3

x = 6 : m = \frac{1}{6}

Find the line with slope m=

\frac{1}{6}

and passing through

(6, 3) : y = \frac{1}{6} x + 2

y = \frac{1}{6} x + 2

y^{^{''}} + 16 y = 6 tan (4 x)

\frac{d y}{d x}, x e^{y} - 10 x + 3 y = 0

(sin (3 x))^{l n (x)}

\frac{\partial}{\partial x} (2 x + y^{2})

\frac{d}{d x} (arcsinh (tan (x)))

What is the tangent of f(x)=sqrt(x+3),\at x=6 ?
The tangent of f(x)=sqrt(x+3),\at x=6 is y= 1/6 x+2