What is the tangent of y^3=126-x^3,\at x=5 ?

The tangent of y^3=126-x^3,\at x=5 is y=-25x+126

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tangent of y^3=126-x^3,\at x=5

tangent of

y^{3} = 126 - x^{3}, at x = 5

y = - 25 x + 126

Solution steps

Find the tangent point:

(5, 1)

Find the slope of

y^{3} = 126 - x^{3} : \frac{d y}{d x} = - \frac{x ^{2}}{( 126 - x ^{3} ) ^{\frac{2}{3}}}

Compute the slope of

y^{3} = 126 - x^{3}

x = 5 : m = - 25

Find the line with slope m=

- 25

and passing through

(5, 1) : y = - 25 x + 126

y = - 25 x + 126

\frac{d}{d x} (2^{3} e^{2})

\int \frac{x ^{2}}{a x ^{2} - b} d x

\frac{d}{d x} (arctan^{2} (6 x))

y^{^{''}} - 12 y^{^{'}} + 100 y = 0

y^{^{'}} = 5 y + e^{- 2 x} y^{- 2}, y (0) = 2

What is the tangent of y^3=126-x^3,\at x=5 ?
The tangent of y^3=126-x^3,\at x=5 is y=-25x+126