What is the tangent of y=(5x-1)(4+3x)x,\at x=0 ?

The tangent of y=(5x-1)(4+3x)x,\at x=0 is y=-4x

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tangent of y=(5x-1)(4+3x)x,\at x=0

tangent of

y = (5 x - 1) (4 + 3 x) x, at x = 0

y = - 4 x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

y = (5 x - 1) (4 + 3 x) x : \frac{d y}{d x} = 45 x^{2} + 34 x - 4

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 = - 4

Find the line with slope m=

- 4

and passing through

(0, 0) : y = - 4 x

y = - 4 x

n = 1 \sum \infty \frac{7}{3 ^{n}}

\frac{d}{d x} (x^{2} + \frac{16}{x})

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 1, y^{^{'}} (0) = 1

\frac{d}{d x} ((x + 1)^{s i n (x)})

2 \frac{d y}{d x} = y - e^{4 x}

What is the tangent of y=(5x-1)(4+3x)x,\at x=0 ?
The tangent of y=(5x-1)(4+3x)x,\at x=0 is y=-4x