What is the tangent of y= 1/(x^3),(1,1) ?

The tangent of y= 1/(x^3),(1,1) is y=-3x+4

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tangent of y= 1/(x^3),(1,1)

tangent of

y = \frac{1}{x ^{3}}, (1, 1)

y = - 3 x + 4

Solution steps

Find the slope of

y = \frac{1}{x ^{3}} : \frac{d y}{d x} = - \frac{3}{x ^{4}}

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 = - 3

Find the line with slope m=

- 3

and passing through

(1, 1) : y = - 3 x + 4

y = - 3 x + 4

f (x) = \frac{5 x}{x - 3}, (4, 20)

y = (x + 5)^{\frac{1}{x}}

\int x^{2} (x^{3} + 5)^{9} d x

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

\int \frac{3 - 6 x}{25 - 9 x ^{2}} d x

What is the tangent of y= 1/(x^3),(1,1) ?
The tangent of y= 1/(x^3),(1,1) is y=-3x+4