What is the tangent of f(x)= 4/(x^5)-1/(x^2),\at x=1 ?

The tangent of f(x)= 4/(x^5)-1/(x^2),\at x=1 is y=-18x+21

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tangent of f(x)= 4/(x^5)-1/(x^2),\at x=1

tangent of

f (x) = \frac{4}{x ^{5}} - \frac{1}{x ^{2}}, at x = 1

y = - 18 x + 21

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

f (x) = \frac{4}{x ^{5}} - \frac{1}{x ^{2}} : \frac{df ( x )}{d x} = - \frac{20}{x ^{6}} + \frac{2}{x ^{3}}

Compute the slope of

f (x) = \frac{4}{x ^{5}} - \frac{1}{x ^{2}}

x = 1 : m = - 18

Find the line with slope m=

- 18

and passing through

(1, 3) : y = - 18 x + 21

y = - 18 x + 21

y = x^{3} - 13 x^{2} + 48 x + 8

\int_{0}^{1} (x)^{2} d x

\frac{\partial}{\partial x} (5 y)

\frac{d}{d x} y = (x^{2} + 3 x)^{2}

\frac{\partial}{\partial x} (6 - x^{7} - y^{5})

What is the tangent of f(x)= 4/(x^5)-1/(x^2),\at x=1 ?
The tangent of f(x)= 4/(x^5)-1/(x^2),\at x=1 is y=-18x+21