What is the tangent of f(x)=-6/((3x+5)^2),\at x=-1 ?

The tangent of f(x)=-6/((3x+5)^2),\at x=-1 is y= 9/2 x+3

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tangent of f(x)=-6/((3x+5)^2),\at x=-1

tangent of

f (x) = - \frac{6}{( 3 x + 5 ) ^{2}}, at x = - 1

y = \frac{9}{2} x + 3

Solution steps

Find the tangent point:

(- 1, - \frac{3}{2})

Find the slope of

f (x) = - \frac{6}{( 3 x + 5 ) ^{2}} : \frac{df ( x )}{d x} = \frac{36}{( 3 x + 5 ) ^{3}}

Compute the slope of

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

x = - 1 : m = \frac{9}{2}

Find the line with slope m=

\frac{9}{2}

and passing through

(- 1, - \frac{3}{2}) : y = \frac{9}{2} x + 3

y = \frac{9}{2} x + 3

f (x) = x^{2} - 1

\int \frac{x ^{3}}{x ^{4} - 2} d x

f (x) = \frac{1}{x - 2}

\frac{\partial}{\partial x} (\frac{- 4 x ^{5}}{4 y ^{4}})

What is the tangent of f(x)=-6/((3x+5)^2),\at x=-1 ?
The tangent of f(x)=-6/((3x+5)^2),\at x=-1 is y= 9/2 x+3