What is the tangent of (4x-3)^{1/2} ?

The tangent of (4x-3)^{1/2} is y= 2/((4a_{0)-3)^{1/2}}x+(4a_{0}-3)^{1/2}-(2a_{0})/((4a_{0)-3)^{1/2}}

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tangent of (4x-3)^{1/2}

tangent of

(4 x - 3)^{\frac{1}{2}}

y = \frac{2}{( 4 a _{0} - 3 ) ^{\frac{1}{2}}} x + (4 a_{0} - 3)^{\frac{1}{2}} - \frac{2 a _{0}}{( 4 a _{0} - 3 ) ^{\frac{1}{2}}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, (4 a_{0} - 3)^{\frac{1}{2}})

Find the slope of

y = (4 x - 3)^{\frac{1}{2}} : \frac{d y}{d x} = \frac{2}{( 4 x - 3 ) ^{\frac{1}{2}}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{2}{( 4 a _{0} - 3 ) ^{\frac{1}{2}}}

Find the line with slope m=

\frac{2}{( 4 a _{0} - 3 ) ^{\frac{1}{2}}}

and passing through

(a_{0}, (4 a_{0} - 3)^{\frac{1}{2}}) : y = \frac{2}{( 4 a _{0} - 3 ) ^{\frac{1}{2}}} x + (4 a_{0} - 3)^{\frac{1}{2}} - \frac{2 a _{0}}{( 4 a _{0} - 3 ) ^{\frac{1}{2}}}

y = \frac{2}{( 4 a _{0} - 3 ) ^{\frac{1}{2}}} x + (4 a_{0} - 3)^{\frac{1}{2}} - \frac{2 a _{0}}{( 4 a _{0} - 3 ) ^{\frac{1}{2}}}

What is the tangent of (4x-3)^{1/2} ?
The tangent of (4x-3)^{1/2} is y= 2/((4a_{0)-3)^{1/2}}x+(4a_{0}-3)^{1/2}-(2a_{0})/((4a_{0)-3)^{1/2}}