What is the tangent of f(x)=(x^2)/(x+9),\at x=1 ?

The tangent of f(x)=(x^2)/(x+9),\at x=1 is y= 19/100 x-9/100

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tangent of f(x)=(x^2)/(x+9),\at x=1

tangent of

f (x) = \frac{x ^{2}}{x + 9}, at x = 1

y = \frac{19}{100} x - \frac{9}{100}

Solution steps

Find the tangent point:

(1, \frac{1}{10})

Find the slope of

f (x) = \frac{x ^{2}}{x + 9} : \frac{df ( x )}{d x} = \frac{x ^{2} + 18 x}{( x + 9 ) ^{2}}

Compute the slope of

f (x) = \frac{x ^{2}}{x + 9}

x = 1 : m = \frac{19}{100}

Find the line with slope m=

\frac{19}{100}

and passing through

(1, \frac{1}{10}) : y = \frac{19}{100} x - \frac{9}{100}

y = \frac{19}{100} x - \frac{9}{100}

x \to 0 lim (\frac{x}{x ^{2} + 5 x})

y = x^{4} f (x)

\int \frac{1}{x ( 1 + ln ( x ) ) ^{4}} d x

\frac{6}{2 s + 1}

\int cot (x) ln (sin (x)) d x

What is the tangent of f(x)=(x^2)/(x+9),\at x=1 ?
The tangent of f(x)=(x^2)/(x+9),\at x=1 is y= 19/100 x-9/100