What is the tangent of f(x)= 3/((x+5)^2),\at x=1 ?

The tangent of f(x)= 3/((x+5)^2),\at x=1 is y=-1/36 x+1/9

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tangent of f(x)= 3/((x+5)^2),\at x=1

tangent of

f (x) = \frac{3}{( x + 5 ) ^{2}}, at x = 1

y = - \frac{1}{36} x + \frac{1}{9}

Solution steps

Find the tangent point:

(1, \frac{1}{12})

Find the slope of

f (x) = \frac{3}{( x + 5 ) ^{2}} : \frac{df ( x )}{d x} = - \frac{6}{( x + 5 ) ^{3}}

Compute the slope of

f (x) = \frac{3}{( x + 5 ) ^{2}}

x = 1 : m = - \frac{1}{36}

Find the line with slope m=

- \frac{1}{36}

and passing through

(1, \frac{1}{12}) : y = - \frac{1}{36} x + \frac{1}{9}

y = - \frac{1}{36} x + \frac{1}{9}

x \to 3 lim (\frac{∣ x + 7 ∣}{x - 3})

f (t) = 3 t - t^{2}, (0, 0)

\int (x + 3)^{4} d x

\frac{d}{d x} (x^{2} sin (y))

y^{^{'}} = 100 - y

What is the tangent of f(x)= 3/((x+5)^2),\at x=1 ?
The tangent of f(x)= 3/((x+5)^2),\at x=1 is y=-1/36 x+1/9