What is the tangent of f(x)=6sqrt(x),\at x=9 ?

The tangent of f(x)=6sqrt(x),\at x=9 is y=x+9

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tangent of f(x)=6sqrt(x),\at x=9

tangent of

f (x) = 6 x, at x = 9

y = x + 9

Solution steps

Find the tangent point:

(9, 18)

Find the slope of

f (x) = 6 x : \frac{df ( x )}{d x} = \frac{3}{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 = 1

Find the line with slope m=

1

and passing through

(9, 18) : y = x + 9

y = x + 9

\int 2 e^{x} x d x

d er i v a t i v e y = ln (x^{5})

\frac{d}{d x} (- \frac{3}{x ^{2}} + 4 e^{2 x})

f (x) = \frac{x ^{2}}{3}

\frac{\partial}{\partial y} (2 x + 3 y - 7)

What is the tangent of f(x)=6sqrt(x),\at x=9 ?
The tangent of f(x)=6sqrt(x),\at x=9 is y=x+9