What is the tangent of 3x^2+8 ?

The tangent of 3x^2+8 is y=6a_{0}x-3a_{0}^2+8

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tangent of 3x^2+8

tangent of

3 x^{2} + 8

y = 6 a_{0} x - 3 a_{0}^{2} + 8

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 3 a_{0}^{2} + 8)

Find the slope of

y = 3 x^{2} + 8 : \frac{d y}{d x} = 6 x

Compute the slope of

y = 3 x^{2} + 8

x = a_{0} : m = 6 a_{0}

Find the line with slope m=

6 a_{0}

and passing through

(a_{0}, 3 a_{0}^{2} + 8) : y = 6 a_{0} x - 3 a_{0}^{2} + 8

y = 6 a_{0} x - 3 a_{0}^{2} + 8

\int \frac{1}{v ^{2} - v} d v

f (θ) = 3 sin (θ)

\frac{\partial}{\partial y} (5 x y^{2} + x^{2})

t^{2} y^{^{''}} + t y^{^{'}} + 64 y = 0

\frac{d}{d v} (\frac{m}{v})