What is the tangent of y=t^3-36t+4,(6,4) ?

The tangent of y=t^3-36t+4,(6,4) is y=72t-428

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tangent of y=t^3-36t+4,(6,4)

tangent of

y = t^{3} - 36 t + 4, (6, 4)

y = 72 t - 428

Solution steps

Find the slope of

y = t^{3} - 36 t + 4 : \frac{d y}{d t} = 3 t^{2} - 36

Compute the slope of

y = t^{3} - 36 t + 4

(6, 4) : m = 72

Find the line with slope m=

72

and passing through

(6, 4) : y = 72 t - 428

y = 72 t - 428

x \to 0 - lim (e^{\frac{4}{x ^{5}}})

\int \frac{x ^{2} - 6}{( x + 2 ) ( 2 x - 1 )} d x

y = x^{2}, y = 6 + x

\frac{d}{d x} (lo g_{e} (4 x))

\frac{d}{d x} ((8 (4 x)^{4} + 5)^{3})

What is the tangent of y=t^3-36t+4,(6,4) ?
The tangent of y=t^3-36t+4,(6,4) is y=72t-428