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

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

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

tangent of

y = x, (25, 5)

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

Solution steps

Find the slope of

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

Compute the slope of

y = 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}

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

\frac{d}{d x} (\frac{1}{π} c x^{2} + ln (cos^{3} (x)))

\int_{1}^{9} \frac{1}{6 - x} d x

\int \frac{1}{x} 1 + x^{2} d x

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