What is the tangent of f(x)=sqrt(5x+39),\at x=5 ?

The tangent of f(x)=sqrt(5x+39),\at x=5 is y= 5/16 x+103/16

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tangent of f(x)=sqrt(5x+39),\at x=5

tangent of

f (x) = 5 x + 39, at x = 5

y = \frac{5}{16} x + \frac{103}{16}

Solution steps

Find the tangent point:

(5, 8)

Find the slope of

f (x) = 5 x + 39 : \frac{df ( x )}{d x} = \frac{5}{2 5 x + 39}

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{5}{16}

Find the line with slope m=

\frac{5}{16}

and passing through

(5, 8) : y = \frac{5}{16} x + \frac{103}{16}

y = \frac{5}{16} x + \frac{103}{16}

\frac{x}{5 + x ^{2}}

\frac{20}{- 5 s + 3}

\frac{d}{d x} (\frac{1}{x} + 2 x)

sin (t - \frac{π}{2})

f (x) = 3 - 4 x

What is the tangent of f(x)=sqrt(5x+39),\at x=5 ?
The tangent of f(x)=sqrt(5x+39),\at x=5 is y= 5/16 x+103/16