What is the tangent of f(x)=sqrt(x),(49,7) ?

The tangent of f(x)=sqrt(x),(49,7) is y= 1/14 x+7/2

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tangent of f(x)=sqrt(x),(49,7)

tangent of

f (x) = x, (49, 7)

y = \frac{1}{14} x + \frac{7}{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

(49, 7) : m = \frac{1}{14}

Find the line with slope m=

\frac{1}{14}

and passing through

(49, 7) : y = \frac{1}{14} x + \frac{7}{2}

y = \frac{1}{14} x + \frac{7}{2}

\frac{4 x}{5 - cot ( x )}

\frac{\partial}{\partial y} (x^{\frac{y}{z}})

\frac{\partial}{\partial x} (x^{3} y^{2} z)

\int \frac{x ^{2} + 2}{x ( x - 1 ) ^{3}} d x

\frac{d}{d x} (- x - \frac{16}{x})

What is the tangent of f(x)=sqrt(x),(49,7) ?
The tangent of f(x)=sqrt(x),(49,7) is y= 1/14 x+7/2