What is the tangent of f(x)= 2/(3x^{2/3)},\at x=64 ?

The tangent of f(x)= 2/(3x^{2/3)},\at x=64 is y=-1/2304 x+5/72

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tangent of f(x)= 2/(3x^{2/3)},\at x=64

tangent of

f (x) = \frac{2}{3 x ^{\frac{2}{3}}}, at x = 64

y = - \frac{1}{2304} x + \frac{5}{72}

Solution steps

Find the tangent point:

(64, \frac{1}{24})

Find the slope of

f (x) = \frac{2}{3 x ^{\frac{2}{3}}} : \frac{df ( x )}{d x} = - \frac{4}{9 x ^{\frac{5}{3}}}

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}{2304}

Find the line with slope m=

- \frac{1}{2304}

and passing through

(64, \frac{1}{24}) : y = - \frac{1}{2304} x + \frac{5}{72}

y = - \frac{1}{2304} x + \frac{5}{72}

\frac{\partial}{\partial x} (64 - 8 x^{2} - y^{2})

x \to 1 + lim (\frac{1 - x}{x ( x - 1 )})

\frac{d}{d x} (cos^{2} (3 π x))

x \to - 9 - lim (\frac{x + 8}{x + 9})

\int \frac{( x ^{2} + 1 )}{x ( x - 1 ) ^{3}} d x

What is the tangent of f(x)= 2/(3x^{2/3)},\at x=64 ?
The tangent of f(x)= 2/(3x^{2/3)},\at x=64 is y=-1/2304 x+5/72