What is the tangent of y=3x^3+5x+4,(2,38) ?

The tangent of y=3x^3+5x+4,(2,38) is y=41x-44

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tangent of y=3x^3+5x+4,(2,38)

tangent of

y = 3 x^{3} + 5 x + 4, (2, 38)

y = 41 x - 44

Solution steps

Find the slope of

y = 3 x^{3} + 5 x + 4 : \frac{d y}{d x} = 9 x^{2} + 5

Compute the slope of

y = 3 x^{3} + 5 x + 4

(2, 38) : m = 41

Find the line with slope m=

41

and passing through

(2, 38) : y = 41 x - 44

y = 41 x - 44

\frac{d y}{d x} = y x + y^{2}

\int \frac{x ^{2} + 5 x}{( x - 3 ) ( x ^{2} + 9 )} d x

y = x^{2} - 37, y = 11 - 2 x

y = e^{- 6 x^{5}}

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

What is the tangent of y=3x^3+5x+4,(2,38) ?
The tangent of y=3x^3+5x+4,(2,38) is y=41x-44