What is the tangent of y=3x^2+2 ?

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

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tangent of y=3x^2+2

tangent of

y = 3 x^{2} + 2

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

y = 3 x^{2} + 2

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

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

\frac{d}{d x} ((\frac{5 ^{x}}{x ^{5}})^{s i n (x^{5})})

x \to - 2 lim (2016)

f (x) = \frac{2 x ^{2} tan ( x )}{sec ( x )}

x \to 1 lim (x + x)

(1 - x)^{^{'}}

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