What is the tangent of f(x)=sqrt(5x+1),\at x=3 ?

The tangent of f(x)=sqrt(5x+1),\at x=3 is y= 5/8 x+17/8

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tangent of f(x)=sqrt(5x+1),\at x=3

tangent of

f (x) = 5 x + 1, at x = 3

y = \frac{5}{8} x + \frac{17}{8}

Solution steps

Find the tangent point:

(3, 4)

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 = 3 : m = \frac{5}{8}

Find the line with slope m=

\frac{5}{8}

and passing through

(3, 4) : y = \frac{5}{8} x + \frac{17}{8}

y = \frac{5}{8} x + \frac{17}{8}

\frac{\partial}{\partial x} (x - 3 y^{2} + 4 x^{2} y^{3})

4 x^{2} + 5

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

\frac{d}{d x} (e^{x} + ln (x))

y^{^{''}} - 4 y^{^{'}} = 6 t

What is the tangent of f(x)=sqrt(5x+1),\at x=3 ?
The tangent of f(x)=sqrt(5x+1),\at x=3 is y= 5/8 x+17/8