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

n = 0 \sum \infty \frac{1}{( n + 1 )}

n = 1 \sum \infty \frac{n !}{n 7 ^{n}}

f (x) = 2 cos (3 x), at x = \frac{π}{2}

y = \frac{1}{2} x^{2}, (- 1, - 1)

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