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)

Compute the slope of

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

(5, 0) : m = 0

Find the line with slope m=

0

and passing through

(5, 0) : y = 0

y = 0

\int x sin (d) x d x

x \to \frac{π}{4} lim (sec (x))

\frac{2}{3 ( s - 3 ) ^{2} + 12}

\int x (x^{2} + 1)^{2} d x

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