What is the tangent of y=5x^2-x^3,(1,4) ?

The tangent of y=5x^2-x^3,(1,4) is y=7x-3

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tangent of y=5x^2-x^3,(1,4)

tangent of

y = 5 x^{2} - x^{3}, (1, 4)

y = 7 x - 3

Solution steps

Find the slope of

y = 5 x^{2} - x^{3} : \frac{d y}{d x} = 10 x - 3 x^{2}

Compute the slope of

y = 5 x^{2} - x^{3}

(1, 4) : m = 7

Find the line with slope m=

7

and passing through

(1, 4) : y = 7 x - 3

y = 7 x - 3

x \to 2 lim (\frac{2}{x})

\int 8 x + cos (x) d x

e^{3 x}, 9

\frac{\partial}{\partial x} (sin (x) cos (4 y))

\int 3 x^{2} - 2 x d x

What is the tangent of y=5x^2-x^3,(1,4) ?
The tangent of y=5x^2-x^3,(1,4) is y=7x-3