What is the tangent of f(x)=sqrt(x),(25,5) ?

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

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tangent of f(x)=sqrt(x),(25,5)

tangent of

f (x) = x, (25, 5)

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

Solution steps

Find the slope of

f (x) = x : \frac{df ( x )}{d x} = \frac{1}{2 x}

Compute the slope of

f (x) = x

(25, 5) : 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}

\int (u - 2)^{3} ln (u - 2) d u

\frac{d}{d x} (\frac{x + 7}{x})

\int \frac{1}{1 - x ^{2} arcsin ( x )} d x

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

\frac{d}{d x} (ln (h + \frac{1}{2 a x ^{2}} + b))

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