What is the tangent of f(x)=sqrt(x),\at x= 49/121 ?

The tangent of f(x)=sqrt(x),\at x= 49/121 is y= 11/14 x+7/22

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tangent of f(x)=sqrt(x),\at x= 49/121

tangent of

f (x) = x, at x = \frac{49}{121}

y = \frac{11}{14} x + \frac{7}{22}

Solution steps

Find the tangent point:

(\frac{49}{121}, \frac{7}{11})

Find the slope of

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

Compute the slope of

f (x) = x

x = \frac{49}{121} : m = \frac{11}{14}

Find the line with slope m=

\frac{11}{14}

and passing through

(\frac{49}{121}, \frac{7}{11}) : y = \frac{11}{14} x + \frac{7}{22}

y = \frac{11}{14} x + \frac{7}{22}

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

\frac{d ^{4}}{d x ^{4}} (2 e^{x^{2}})

- 3 y

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

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

What is the tangent of f(x)=sqrt(x),\at x= 49/121 ?
The tangent of f(x)=sqrt(x),\at x= 49/121 is y= 11/14 x+7/22