What is the tangent of y=(e^{2x}-2)^2,(0,1) ?

The tangent of y=(e^{2x}-2)^2,(0,1) is y=-4x+1

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tangent of y=(e^{2x}-2)^2,(0,1)

tangent of

y = (e^{2 x} - 2)^{2}, (0, 1)

y = - 4 x + 1

Solution steps

Find the slope of

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

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, 1) : y = - 4 x + 1

y = - 4 x + 1

\frac{d y}{d x}, \frac{y}{x} = 0

\frac{\partial}{\partial x} (sin (h) (x^{2} + y))

\frac{d}{d x} (- ln (x^{2} + y))

x \to 7 lim (x)

\frac{1}{s ^{2} + s + 1}

What is the tangent of y=(e^{2x}-2)^2,(0,1) ?
The tangent of y=(e^{2x}-2)^2,(0,1) is y=-4x+1