What is the tangent of f(x)= 1/(sqrt(x+1)),(0,1) ?

The tangent of f(x)= 1/(sqrt(x+1)),(0,1) is y=-1/2 x+1

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tangent of f(x)= 1/(sqrt(x+1)),(0,1)

tangent of

f (x) = \frac{1}{x + 1}, (0, 1)

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

Solution steps

Find the slope of

f (x) = \frac{1}{x + 1} : \frac{df ( x )}{d x} = - \frac{1}{2 ( x + 1 ) ^{\frac{3}{2}}}

Compute the slope of

f (x) = \frac{1}{x + 1}

(0, 1) : m = - \frac{1}{2}

Find the line with slope m=

- \frac{1}{2}

and passing through

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

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

y = 6 e^{x}, y = 6 e^{- x}, x = 1

\int \frac{1}{y ^{2} - 9} d y

\int \frac{1}{9 - 64 x ^{2}} d x

y (θ) = π^{4} cos (θ)

\frac{d}{d t} (9 + t)

What is the tangent of f(x)= 1/(sqrt(x+1)),(0,1) ?
The tangent of f(x)= 1/(sqrt(x+1)),(0,1) is y=-1/2 x+1