What is the tangent of f(x)= 6/(sqrt(x)),\at x=36 ?

The tangent of f(x)= 6/(sqrt(x)),\at x=36 is y=-1/72 x+3/2

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tangent of f(x)= 6/(sqrt(x)),\at x=36

tangent of

f (x) = \frac{6}{x}, at x = 36

y = - \frac{1}{72} x + \frac{3}{2}

Solution steps

Find the tangent point:

(36, 1)

Find the slope of

f (x) = \frac{6}{x} : \frac{df ( x )}{d x} = - \frac{3}{x ^{\frac{3}{2}}}

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}{72}

Find the line with slope m=

- \frac{1}{72}

and passing through

(36, 1) : y = - \frac{1}{72} x + \frac{3}{2}

y = - \frac{1}{72} x + \frac{3}{2}

x \to 2 lim (\frac{x - 2}{x ^{2} - 4})

\frac{d}{d x} (\frac{cos ( 12 x )}{e ^{7 x}})

\int \frac{sec ^{6} ( x )}{tan ^{2} ( x )} d x

d er i v a t i v e 9 t - 2 t^{2}

\int 4 x^{- 2} d x

What is the tangent of f(x)= 6/(sqrt(x)),\at x=36 ?
The tangent of f(x)= 6/(sqrt(x)),\at x=36 is y=-1/72 x+3/2