What is the tangent of f(x)= 1/(x^5),(2, 1/32) ?

The tangent of f(x)= 1/(x^5),(2, 1/32) is y=-5/64 x+3/16

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tangent of f(x)= 1/(x^5),(2, 1/32)

tangent of

f (x) = \frac{1}{x ^{5}}, (2, \frac{1}{32})

y = - \frac{5}{64} x + \frac{3}{16}

Solution steps

Find the slope of

f (x) = \frac{1}{x ^{5}} : \frac{df ( x )}{d x} = - \frac{5}{x ^{6}}

Compute the slope of

f (x) = \frac{1}{x ^{5}}

(2, \frac{1}{32}) : m = - \frac{5}{64}

Find the line with slope m=

- \frac{5}{64}

and passing through

(2, \frac{1}{32}) : y = - \frac{5}{64} x + \frac{3}{16}

y = - \frac{5}{64} x + \frac{3}{16}

f (t) = 2 t^{2} - t

y^{^{'}} + \frac{1}{x + 1} y = x^{- 2}

f (x) = (3 x - 2) (x + 4), (1, 5)

\frac{\partial}{\partial y} (x \cdot cos (y) \cdot sin (z))

x \to \infty lim (\frac{- x}{x})

What is the tangent of f(x)= 1/(x^5),(2, 1/32) ?
The tangent of f(x)= 1/(x^5),(2, 1/32) is y=-5/64 x+3/16