What is the tangent of 4/(sqrt(x-7)) ?

The tangent of 4/(sqrt(x-7)) is y=-2/((a_{0)-7)^{3/2}}x+(6a_{0}-28)/((a_{0)-7)^{3/2}}

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tangent of 4/(sqrt(x-7))

tangent of

\frac{4}{x - 7}

y = - \frac{2}{( a _{0} - 7 ) ^{\frac{3}{2}}} x + \frac{6 a _{0} - 28}{( a _{0} - 7 ) ^{\frac{3}{2}}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = \frac{4}{x - 7} : \frac{d y}{d x} = - \frac{2}{( x - 7 ) ^{\frac{3}{2}}}

Compute the slope of

y = \frac{4}{x - 7}

x = a_{0} : m = - \frac{2}{( a _{0} - 7 ) ^{\frac{3}{2}}}

Find the line with slope m=

- \frac{2}{( a _{0} - 7 ) ^{\frac{3}{2}}}

and passing through

(a_{0}, \frac{4}{a _{0} - 7}) : y = - \frac{2}{( a _{0} - 7 ) ^{\frac{3}{2}}} x + \frac{6 a _{0} - 28}{( a _{0} - 7 ) ^{\frac{3}{2}}}

y = - \frac{2}{( a _{0} - 7 ) ^{\frac{3}{2}}} x + \frac{6 a _{0} - 28}{( a _{0} - 7 ) ^{\frac{3}{2}}}

\frac{\partial}{\partial x} (x^{2} + 2 x y - y^{2} - 8 x)

f (t) = - 4 t^{2} + 16 t + 9

\frac{d}{d x} ((3 x^{2} - 4 x) e^{x})

\int \frac{4 x ^{3} - 7 x - 12}{x ^{3} - x ^{2} - 6 x} d x

\int 15 sin^{2} (x) cos^{3} (x) d x

What is the tangent of 4/(sqrt(x-7)) ?
The tangent of 4/(sqrt(x-7)) is y=-2/((a_{0)-7)^{3/2}}x+(6a_{0}-28)/((a_{0)-7)^{3/2}}