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

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

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

tangent of

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

y = \frac{1}{2 6} x + 6 - \frac{5}{2 6}

Solution steps

Find the tangent point:

(5, 6)

Find the slope of

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

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

(5, 6) : y = \frac{1}{2 6} x + 6 - \frac{5}{2 6}

y = \frac{1}{2 6} x + 6 - \frac{5}{2 6}

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

\int_{0}^{\frac{π}{6}} 2 sec^{3} (x) d x

\int 1 - e^{- x} d x

\int_{1}^{7} \frac{x + 1}{x ^{2} + 2 x - 1} d x

y = \frac{π ^{2} x ^{6} + 9000}{x ^{2}}

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