What is the tangent of f(x)=sqrt(x+5),(4,3) ?

The tangent of f(x)=sqrt(x+5),(4,3) is y= 1/6 x+7/3

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tangent of f(x)=sqrt(x+5),(4,3)

tangent of

f (x) = x + 5, (4, 3)

y = \frac{1}{6} x + \frac{7}{3}

Solution steps

Find the slope of

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

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{1}{6}

Find the line with slope m=

\frac{1}{6}

and passing through

(4, 3) : y = \frac{1}{6} x + \frac{7}{3}

y = \frac{1}{6} x + \frac{7}{3}

\frac{d}{d x} (25 - x^{2} - y^{2})

f (x) = \frac{x ^{2}}{x + 8}, (1, \frac{1}{9})

y^{^{'}} = y^{2} + 1

f (x) = x 4 - x

\frac{d}{d x} (2 x + 3 y)

What is the tangent of f(x)=sqrt(x+5),(4,3) ?
The tangent of f(x)=sqrt(x+5),(4,3) is y= 1/6 x+7/3