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

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

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

tangent of

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

y = 5 x - 2

Solution steps

Find the slope of

y = 4 x^{2} - x^{3} : \frac{d y}{d x} = 8 x - 3 x^{2}

Compute the slope of

y = 4 x^{2} - x^{3}

(1, 3) : m = 5

Find the line with slope m=

5

and passing through

(1, 3) : y = 5 x - 2

y = 5 x - 2

e^{x}, e^{2 x}, ln (2), ln (4)

\frac{\partial}{\partial x} (8 - 3 x^{2} - y^{2})

\int \frac{x ^{2} + 2 x - 6}{x ^{3} - x} d x

\frac{d}{d x} (e^{x \cdot l n (x)})

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

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