What is the tangent of y=(x^3-25x)^{12},(-5,0) ?

The tangent of y=(x^3-25x)^{12},(-5,0) is y=0

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tangent of y=(x^3-25x)^{12},(-5,0)

tangent of

y = (x^{3} - 25 x)^{12}, (- 5, 0)

y = 0

Solution steps

Find the slope of

y = (x^{3} - 25 x)^{12} : \frac{d y}{d x} = 12 (x^{3} - 25 x)^{11} (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

\int a + 18

\frac{d}{d x} (sin (7 π x))

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

\int_{0}^{\frac{π}{24}} cos (4 x) d x

\int 9 + 16 t^{2} d t

What is the tangent of y=(x^3-25x)^{12},(-5,0) ?
The tangent of y=(x^3-25x)^{12},(-5,0) is y=0