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

The tangent of f(x)=(x/(1+x))^5,\at x=1 is y= 5/64 x-3/64

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

tangent of

f (x) = (\frac{x}{1 + x})^{5}, at x = 1

y = \frac{5}{64} x - \frac{3}{64}

Solution steps

Find the tangent point:

(1, \frac{1}{32})

Find the slope of

f (x) = (\frac{x}{1 + x})^{5} : \frac{df ( x )}{d x} = \frac{5 x ^{4}}{( 1 + x ) ^{6}}

Compute the slope of

f (x) = (\frac{x}{1 + x})^{5}

x = 1 : m = \frac{5}{64}

Find the line with slope m=

\frac{5}{64}

and passing through

(1, \frac{1}{32}) : y = \frac{5}{64} x - \frac{3}{64}

y = \frac{5}{64} x - \frac{3}{64}

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What is the tangent of f(x)=(x/(1+x))^5,\at x=1 ?
The tangent of f(x)=(x/(1+x))^5,\at x=1 is y= 5/64 x-3/64