What is the tangent of y=3x^2+2,\at x=1 ?

The tangent of y=3x^2+2,\at x=1 is y=6x-1

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tangent of y=3x^2+2,\at x=1

tangent of

y = 3 x^{2} + 2, at x = 1

y = 6 x - 1

Solution steps

Find the tangent point:

(1, 5)

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 6

Find the line with slope m=

6

and passing through

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

y = 6 x - 1

\int x tan (x^{2}) d x

\frac{d y}{d x} = 2 x y + 3 y - 4 x - 6

\frac{x d y}{d x} = 6 y

x \to 7 lim (x^{2} - 49)

\int \frac{8 + x ^{2}}{x} d x

What is the tangent of y=3x^2+2,\at x=1 ?
The tangent of y=3x^2+2,\at x=1 is y=6x-1