What is the tangent of f(x)=(3x)/(1+x^2),\at x=4 ?

The tangent of f(x)=(3x)/(1+x^2),\at x=4 is y=-45/289 x+384/289

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tangent of f(x)=(3x)/(1+x^2),\at x=4

tangent of

f (x) = \frac{3 x}{1 + x ^{2}}, at x = 4

y = - \frac{45}{289} x + \frac{384}{289}

Solution steps

Find the tangent point:

(4, \frac{12}{17})

Find the slope of

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

Compute the slope of

f (x) = \frac{3 x}{1 + x ^{2}}

x = 4 : m = - \frac{45}{289}

Find the line with slope m=

- \frac{45}{289}

and passing through

(4, \frac{12}{17}) : y = - \frac{45}{289} x + \frac{384}{289}

y = - \frac{45}{289} x + \frac{384}{289}

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\frac{d}{d x} (2 5^{x})

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\frac{d}{d x} (8 x - 2)

What is the tangent of f(x)=(3x)/(1+x^2),\at x=4 ?
The tangent of f(x)=(3x)/(1+x^2),\at x=4 is y=-45/289 x+384/289