What is the tangent of f(x)=6sqrt(x),\at x=1 ?

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

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tangent of f(x)=6sqrt(x),\at x=1

tangent of

f (x) = 6 x, at x = 1

y = 3 x + 3

Solution steps

Find the tangent point:

(1, 6)

Find the slope of

f (x) = 6 x : \frac{df ( x )}{d x} = \frac{3}{x}

Compute the slope of

f (x) = 6 x

x = 1 : m = 3

Find the line with slope m=

3

and passing through

(1, 6) : y = 3 x + 3

y = 3 x + 3

(16 y^{2} - x y) d x + x^{2} d y = 0

\frac{d}{d x} (2^{x} tan (3 x))

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

\frac{\partial}{\partial x} (5 x y - 18 x)

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