What is the tangent of y=(x^3-25x)^{14},(5,0) ?

The tangent of y=(x^3-25x)^{14},(5,0) is y=0

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tangent of y=(x^3-25x)^{14},(5,0)

tangent of

y = (x^{3} - 25 x)^{14}, (5, 0)

y = 0

Solution steps

Find the slope of

y = (x^{3} - 25 x)^{14} : \frac{d y}{d x} = 14 (x^{3} - 25 x)^{13} (3 x^{2} - 25)

Compute the slope of

y = (x^{3} - 25 x)^{14}

(5, 0) : m = 0

Find the line with slope m=

0

and passing through

(5, 0) : y = 0

y = 0

\int θ arcsec (4 θ) d θ

y^{^{'}} + 10 y = 2 cos (t)

x y^{^{'}} = 2 y + x^{3} e^{x}

t \to 1 lim (3 t^{3} + 2)

\int \frac{x}{1 + x} d x

What is the tangent of y=(x^3-25x)^{14},(5,0) ?
The tangent of y=(x^3-25x)^{14},(5,0) is y=0