What is the tangent of f(x)= x/(sqrt(x^2+9)),\at x=4 ?

The tangent of f(x)= x/(sqrt(x^2+9)),\at x=4 is y= 9/125 x+64/125

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tangent of f(x)= x/(sqrt(x^2+9)),\at x=4

tangent of

f (x) = \frac{x}{x ^{2} + 9}, at x = 4

y = \frac{9}{125} x + \frac{64}{125}

Solution steps

Find the tangent point:

(4, \frac{4}{5})

Find the slope of

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

Compute the slope of

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

x = 4 : m = \frac{9}{125}

Find the line with slope m=

\frac{9}{125}

and passing through

(4, \frac{4}{5}) : y = \frac{9}{125} x + \frac{64}{125}

y = \frac{9}{125} x + \frac{64}{125}

f (x) = 2 x^{3}, (2, 16)

x \to 3 + lim (\frac{x ^{2} + 1}{x - 3})

(- 8, - 2), (1, 4)

\frac{\partial}{\partial x} (9 y cos (x y))

\frac{\partial}{\partial x} ((x^{2} + x - y)^{7})

What is the tangent of f(x)= x/(sqrt(x^2+9)),\at x=4 ?
The tangent of f(x)= x/(sqrt(x^2+9)),\at x=4 is y= 9/125 x+64/125