What is the tangent of y=(x^3-25x)^8,(-5,0) ?

The tangent of y=(x^3-25x)^8,(-5,0) is y=0

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tangent of y=(x^3-25x)^8,(-5,0)

tangent of

y = (x^{3} - 25 x)^{8}, (- 5, 0)

y = 0

Solution steps

Find the slope of

y = (x^{3} - 25 x)^{8} : \frac{d y}{d x} = 8 (x^{3} - 25 x)^{7} (3 x^{2} - 25)

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 = 0

Find the line with slope m=

0

and passing through

(- 5, 0) : y = 0

y = 0

\frac{d}{d x} (\frac{x ^{4} - 3 x ^{2}}{( x ^{2} - 1 ) ^{2}})

a re a x = \frac{( y - 2 ) ^{2}}{2}, y = 6 - x

t an g e n t f (x) = \frac{x}{( 3 x - 2 ) ^{6}}, at x = 1

\int x^{5} e^{x^{3} - 7} d x

n = 1 \sum \infty \frac{3}{4 n ^{2} - 1}

What is the tangent of y=(x^3-25x)^8,(-5,0) ?
The tangent of y=(x^3-25x)^8,(-5,0) is y=0