What is the tangent of y=(f(x))^2,\at x=0 ?

The tangent of y=(f(x))^2,\at x=0 is y=2f(0) d/(d0)(f(0))x+f(0)^2

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tangent of y=(f(x))^2,\at x=0

tangent of

y = (f (x))^{2}, at x = 0

y = 2 f (0) \frac{d}{d 0} (f (0)) x + f (0)^{2}

Solution steps

Find the tangent point:

(0, f (0)^{2})

Find the slope of

y = f (x)^{2} : \frac{d y}{d x} = 2 f (x) \frac{df}{d 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 = 2 f (0) \frac{d}{d 0} (f (0))

Find the line with slope m=

2 f (0) \frac{d}{d 0} (f (0))

and passing through

(0, f (0)^{2}) : y = 2 f (0) \frac{d}{d 0} (f (0)) x + f (0)^{2}

y = 2 f (0) \frac{d}{d 0} (f (0)) x + f (0)^{2}

What is the tangent of y=(f(x))^2,\at x=0 ?
The tangent of y=(f(x))^2,\at x=0 is y=2f(0) d/(d0)(f(0))x+f(0)^2