What is the tangent of 4x^2,\at x=3 ?

The tangent of 4x^2,\at x=3 is y=24x-36

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tangent of 4x^2,\at x=3

tangent of

4 x^{2}, at x = 3

y = 24 x - 36

Solution steps

Find the tangent point:

(3, 36)

Find the slope of

y = 4 x^{2} : \frac{d y}{d x} = 8 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 = 24

Find the line with slope m=

24

and passing through

(3, 36) : y = 24 x - 36

y = 24 x - 36

\frac{\partial}{\partial θ} (θ)

\int 12 x^{2} - sin (x) d x

x^{2} y^{^{'}} - 2 x^{3} e^{\frac{- 1}{x}} = y

\int \frac{4}{x ( x ^{2} + 4 ) ^{2}} d x

\frac{d}{d x} ((x - 2)^{2} (4 x - 2))