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Popular >

tangent of y=x^3,(2,8)

Solution

tangent of

y = x^{3}, (2, 8)

Solution

y = 12 x - 16

Solution steps

Find the slope of

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

Compute the slope of

y = x^{3}

at

(2, 8) : m = 12

Find the line with slope m=

12

and passing through

(2, 8) : y = 12 x - 16

y = 12 x - 16

Popular Examples

y^{''}+8y^'+16y=2e^{-4t}

y^{^{''}} + 8 y^{^{'}} + 16 y = 2 e^{- 4 t}

(\partial)/(\partial x)(3xln(xy))

\frac{\partial}{\partial x} (3 x ln (x y))

(\partial)/(\partial x)((x^3+x^4)(x^2+3x)^2)

\frac{\partial}{\partial x} ((x^{3} + x^{4}) (x^{2} + 3 x)^{2})

integral from 1 to 4 of (5x^3-6)

\int_{1}^{4} (5 x^{3} - 6) d x

(dy)/(dx)=sqrt(5y)e^{x+8}

\frac{d y}{d x} = 5 y e^{x + 8}

Frequently Asked Questions (FAQ)

What is the tangent of y=x^3,(2,8) ?
The tangent of y=x^3,(2,8) is y=12x-16