What is the tangent of f(x)= 7/(sqrt(x),\at (4, 7/2)) ?

The tangent of f(x)= 7/(sqrt(x),\at (4, 7/2)) is

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tangent of f(x)= 7/(sqrt(x),\at (4, 7/2))

tangent of

f (x) = \frac{7}{x , at ( 4 , \frac{7}{2} )}

y = \frac{7}{, ( 4 , \frac{7}{2} )} (- \frac{1}{2} a_{0}^{- \frac{1}{2} - 1}) x + \frac{7}{a _{0} , ( 4 , \frac{7}{2} )} - \frac{7}{, ( 4 , \frac{7}{2} )} (- \frac{1}{2} a_{0}^{- \frac{1}{2} - 1}) a_{0}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{7}{a _{0} , ( 4 , \frac{7}{2} )})

Find the slope of

f (x) = \frac{7}{x , ( 4 , \frac{7}{2} )} : \frac{df ( x )}{d x} = \frac{7}{, ( 4 , \frac{7}{2} )} (- \frac{1}{2} x^{- \frac{1}{2} - 1})

Compute the slope of

f (x) = \frac{7}{x , ( 4 , \frac{7}{2} )}

x = a_{0} : m = \frac{7}{, ( 4 , \frac{7}{2} )} (- \frac{1}{2} a_{0}^{- \frac{1}{2} - 1})

Find the line with slope m=

\frac{7}{, ( 4 , \frac{7}{2} )} (- \frac{1}{2} a_{0}^{- \frac{1}{2} - 1})

and passing through

(a_{0}, \frac{7}{a _{0} , ( 4 , \frac{7}{2} )}) : y = \frac{7}{, ( 4 , \frac{7}{2} )} (- \frac{1}{2} a_{0}^{- \frac{1}{2} - 1}) x + \frac{7}{a _{0} , ( 4 , \frac{7}{2} )} - \frac{7}{, ( 4 , \frac{7}{2} )} (- \frac{1}{2} a_{0}^{- \frac{1}{2} - 1}) a_{0}

y = \frac{7}{, ( 4 , \frac{7}{2} )} (- \frac{1}{2} a_{0}^{- \frac{1}{2} - 1}) x + \frac{7}{a _{0} , ( 4 , \frac{7}{2} )} - \frac{7}{, ( 4 , \frac{7}{2} )} (- \frac{1}{2} a_{0}^{- \frac{1}{2} - 1}) a_{0}

f (x) = - x^{3}

x^{2} \frac{d y}{d x} + (x^{2} + 2 x) y = e^{x}

x^{2} y^{^{'}} + y^{2} = x y

\frac{d}{d x} ((2 x + 5)^{\frac{1}{3}})

What is the tangent of f(x)= 7/(sqrt(x),\at (4, 7/2)) ?
The tangent of f(x)= 7/(sqrt(x),\at (4, 7/2)) is