What is the tangent of 3x^2+6 ?

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

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tangent of 3x^2+6

tangent of

3 x^{2} + 6

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 3 a_{0}^{2} + 6)

Find the slope of

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

Compute the slope of

y = 3 x^{2} + 6

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

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

\int_{6}^{8} \frac{16}{( x - 6 ) ^{3}} d x

\frac{d}{d x} (x (2 x + 5)^{\frac{1}{3}})

f (x) = 2 x^{3} + 6 x^{2} - 49 x + 24

\frac{\partial}{\partial x} (z arctan (\frac{3 y}{x}))

\int_{- 2.5}^{0} e^{3 x} d x