What is the tangent of sqrt(x),\at x=25 ?

The tangent of sqrt(x),\at x=25 is y= 1/10 x+5/2

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tangent of sqrt(x),\at x=25

tangent of

x, at x = 25

y = \frac{1}{10} x + \frac{5}{2}

Solution steps

Find the tangent point:

(25, 5)

Find the slope of

y = x : \frac{d y}{d x} = \frac{1}{2 x}

Compute the slope of

y = x

x = 25 : m = \frac{1}{10}

Find the line with slope m=

\frac{1}{10}

and passing through

(25, 5) : y = \frac{1}{10} x + \frac{5}{2}

y = \frac{1}{10} x + \frac{5}{2}

\frac{\partial}{\partial x} (sin (- x) cos (4 y))

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

\frac{55}{0.01 x ^{2} + 1}

\int tan^{4} (4 t) d t

\frac{\partial}{\partial y} (x y + ln (x + y))

What is the tangent of sqrt(x),\at x=25 ?
The tangent of sqrt(x),\at x=25 is y= 1/10 x+5/2