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}

Compute the slope of

y = f (x)^{2}

x = 0 : 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}

4500 (1 - \frac{1}{50} t)^{2}

\int \frac{2 x ^{2} + 8 x - 8}{x - 4} d x

f (x) = \frac{( x - 2 ) ^{5}}{16}

x \to \infty lim (\frac{2 x ^{3} - 5 x + 4}{x ^{2} + 3 x - 4})

x cos (x)

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