What is the tangent of f(x)=(x^2)/2 ,\at x=5 ?

The tangent of f(x)=(x^2)/2 ,\at x=5 is y=5x-25/2

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tangent of f(x)=(x^2)/2 ,\at x=5

tangent of

f (x) = \frac{x ^{2}}{2}, at x = 5

y = 5 x - \frac{25}{2}

Solution steps

Find the tangent point:

(5, \frac{25}{2})

Find the slope of

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

Compute the slope of

f (x) = \frac{x ^{2}}{2}

x = 5 : m = 5

Find the line with slope m=

5

and passing through

(5, \frac{25}{2}) : y = 5 x - \frac{25}{2}

y = 5 x - \frac{25}{2}

\frac{d y}{d x} x^{2} y + x y^{2} = 6

e^{- 0.1 s}

\frac{d}{d x} (125 sin (5 x))

x \to 0 lim (x e^{c o s (\frac{1}{x})})

x^{2} - 2, x

What is the tangent of f(x)=(x^2)/2 ,\at x=5 ?
The tangent of f(x)=(x^2)/2 ,\at x=5 is y=5x-25/2