What is the tangent of f(x)=sqrt(x+35),(1,6) ?

The tangent of f(x)=sqrt(x+35),(1,6) is y= 1/12 x+71/12

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tangent of f(x)=sqrt(x+35),(1,6)

tangent of

f (x) = x + 35, (1, 6)

y = \frac{1}{12} x + \frac{71}{12}

Solution steps

Find the slope of

f (x) = x + 35 : \frac{df ( x )}{d x} = \frac{1}{2 x + 35}

Compute the slope of

f (x) = x + 35

(1, 6) : m = \frac{1}{12}

Find the line with slope m=

\frac{1}{12}

and passing through

(1, 6) : y = \frac{1}{12} x + \frac{71}{12}

y = \frac{1}{12} x + \frac{71}{12}

\frac{d y}{d x} + x^{- 1} y = x y^{2}

e^{- t} cos (4 t)

\int (3 x^{2} + 2 x) e^{(x^{3} + x^{2})} d x

f (x) = - x^{3} - 4 x - 3, at x = - 1

\frac{1}{( s - 2 ) ( s - 2 )}

What is the tangent of f(x)=sqrt(x+35),(1,6) ?
The tangent of f(x)=sqrt(x+35),(1,6) is y= 1/12 x+71/12