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

The tangent of sqrt(x+5),\at x=1 is y= 1/(2sqrt(6))x+sqrt(6)-1/(2sqrt(6))

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

tangent of

x + 5, at x = 1

y = \frac{1}{2 6} x + 6 - \frac{1}{2 6}

Solution steps

Find the tangent point:

(1, 6)

Find the slope of

y = x + 5 : \frac{d y}{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}{2 6}

Find the line with slope m=

\frac{1}{2 6}

and passing through

(1, 6) : y = \frac{1}{2 6} x + 6 - \frac{1}{2 6}

y = \frac{1}{2 6} x + 6 - \frac{1}{2 6}

\int \frac{3}{1 - 4 x} d x

\int e^{t a n (9 x)} sec^{2} (9 x) d x

f (x) = x^{2} + 15 x + 56, at x = 2

6 x, x ((2 1^{2} - x^{2}))

y = e^{x} sin (x), (0, 0)

What is the tangent of sqrt(x+5),\at x=1 ?
The tangent of sqrt(x+5),\at x=1 is y= 1/(2sqrt(6))x+sqrt(6)-1/(2sqrt(6))