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

The tangent of 5/(x^2),\at x=1 is y=-10x+15

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

tangent of

\frac{5}{x ^{2}}, at x = 1

y = - 10 x + 15

Solution steps

Find the tangent point:

(1, 5)

Find the slope of

y = \frac{5}{x ^{2}} : \frac{d y}{d x} = - \frac{10}{x ^{3}}

Compute the slope of

y = \frac{5}{x ^{2}}

x = 1 : m = - 10

Find the line with slope m=

- 10

and passing through

(1, 5) : y = - 10 x + 15

y = - 10 x + 15

\frac{t ^{4}}{24}

\int \frac{1}{6 x lo g _{2} ( x )} d x

\int 4 x cos (2 - 3 x) d x

\int (8 x - 5) d x

(cos (2 t))^{^{'}}

What is the tangent of 5/(x^2),\at x=1 ?
The tangent of 5/(x^2),\at x=1 is y=-10x+15