What is the tangent of f(x)= 1/(x^4),(-2, 1/16) ?

The tangent of f(x)= 1/(x^4),(-2, 1/16) is y= 1/8 x+5/16

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tangent of f(x)= 1/(x^4),(-2, 1/16)

tangent of

f (x) = \frac{1}{x ^{4}}, (- 2, \frac{1}{16})

y = \frac{1}{8} x + \frac{5}{16}

Solution steps

Find the slope of

f (x) = \frac{1}{x ^{4}} : \frac{df ( x )}{d x} = - \frac{4}{x ^{5}}

Compute the slope of

f (x) = \frac{1}{x ^{4}}

(- 2, \frac{1}{16}) : m = \frac{1}{8}

Find the line with slope m=

\frac{1}{8}

and passing through

(- 2, \frac{1}{16}) : y = \frac{1}{8} x + \frac{5}{16}

y = \frac{1}{8} x + \frac{5}{16}

\frac{d}{d x} (\frac{c ^{2} - x ^{2}}{2 x})

\int π (\frac{1}{1 + x ^{2}})^{2} d x

\int_{0}^{1} \frac{2 t}{1 + t ^{2}} d t

y = sin (4 x)

\frac{d}{d x} (\frac{10 x + x ^{2}}{( 5 + x ) ^{2}})

What is the tangent of f(x)= 1/(x^4),(-2, 1/16) ?
The tangent of f(x)= 1/(x^4),(-2, 1/16) is y= 1/8 x+5/16