What is the tangent of f(x)=sqrt(5x-1) ?

The tangent of f(x)=sqrt(5x-1) is y= 5/(2sqrt(5a_{0)-1)}x+sqrt(5a_{0)-1}-(5a_{0})/(2sqrt(5a_{0)-1)}

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tangent of f(x)=sqrt(5x-1)

tangent of

f (x) = 5 x - 1

y = \frac{5}{2 5 a _{0} - 1} x + 5 a_{0} - 1 - \frac{5 a _{0}}{2 5 a _{0} - 1}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 5 a_{0} - 1)

Find the slope of

f (x) = 5 x - 1 : \frac{df ( x )}{d x} = \frac{5}{2 5 x - 1}

Compute the slope of

f (x) = 5 x - 1

x = a_{0} : m = \frac{5}{2 5 a _{0} - 1}

Find the line with slope m=

\frac{5}{2 5 a _{0} - 1}

and passing through

(a_{0}, 5 a_{0} - 1) : y = \frac{5}{2 5 a _{0} - 1} x + 5 a_{0} - 1 - \frac{5 a _{0}}{2 5 a _{0} - 1}

y = \frac{5}{2 5 a _{0} - 1} x + 5 a_{0} - 1 - \frac{5 a _{0}}{2 5 a _{0} - 1}

\int_{- 0.5}^{0} 1.5 x^{2} d x

x^{2} y^{^{''}} - 6 x y^{^{'}} + 10 y = 0

4 x^{3} - 4 x

\frac{d}{d x} (x^{2} - 8 ln (x))

\int e^{x} d x

What is the tangent of f(x)=sqrt(5x-1) ?
The tangent of f(x)=sqrt(5x-1) is y= 5/(2sqrt(5a_{0)-1)}x+sqrt(5a_{0)-1}-(5a_{0})/(2sqrt(5a_{0)-1)}