What is the tangent of y=7x^2-x^3,(1,6) ?

The tangent of y=7x^2-x^3,(1,6) is y=11x-5

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tangent of y=7x^2-x^3,(1,6)

tangent of

y = 7 x^{2} - x^{3}, (1, 6)

y = 11 x - 5

Solution steps

Find the slope of

y = 7 x^{2} - x^{3} : \frac{d y}{d x} = 14 x - 3 x^{2}

Compute the slope of

y = 7 x^{2} - x^{3}

(1, 6) : m = 11

Find the line with slope m=

11

and passing through

(1, 6) : y = 11 x - 5

y = 11 x - 5

\int e^{x z} d y

\frac{d}{d x} ((x^{2} - 2)^{3})

\int 5 x^{- 1} d x

\frac{d}{d x} (\frac{1 + x}{x ^{3}})

What is the tangent of y=7x^2-x^3,(1,6) ?
The tangent of y=7x^2-x^3,(1,6) is y=11x-5