What is the derivative of f(x)=((x+3)(x+1))/(x^2-49) ?

The derivative of f(x)=((x+3)(x+1))/(x^2-49) is (-4x^2-104x-196)/((x^2-49)^2)

What is the first derivative of f(x)=((x+3)(x+1))/(x^2-49) ?

The first derivative of f(x)=((x+3)(x+1))/(x^2-49) is (-4x^2-104x-196)/((x^2-49)^2)

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derivative of

f (x) = \frac{( x + 3 ) ( x + 1 )}{x ^{2} - 49}

\frac{- 4 x ^{2} - 104 x - 196}{( x ^{2} - 49 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{( x + 3 ) ( x + 1 )}{x ^{2} - 49})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( ( x + 3 ) ( x + 1 ) ) ( x ^{2} - 49 ) - \frac{d}{d x} ( x ^{2} - 49 ) ( x + 3 ) ( x + 1 )}{( x ^{2} - 49 ) ^{2}}

\frac{d}{d x} ((x + 3) (x + 1)) = 2 x + 4

\frac{d}{d x} (x^{2} - 49) = 2 x

= \frac{( 2 x + 4 ) ( x ^{2} - 49 ) - 2 x ( x + 3 ) ( x + 1 )}{( x ^{2} - 49 ) ^{2}}

Expand

(2 x + 4) (x^{2} - 49) - 2 x (x + 3) (x + 1) : - 4 x^{2} - 104 x - 196

= \frac{- 4 x ^{2} - 104 x - 196}{( x ^{2} - 49 ) ^{2}}

y = ln (x + x^{2} + 6)

x^{\frac{1}{4}}, 16

\frac{d}{d x} (\frac{ln ( - 2 x ^{3} + 5 x ^{2} )}{- x})

f (x) = \frac{1}{1 + x}

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

What is the derivative of f(x)=((x+3)(x+1))/(x^2-49) ?
The derivative of f(x)=((x+3)(x+1))/(x^2-49) is (-4x^2-104x-196)/((x^2-49)^2)
What is the first derivative of f(x)=((x+3)(x+1))/(x^2-49) ?
The first derivative of f(x)=((x+3)(x+1))/(x^2-49) is (-4x^2-104x-196)/((x^2-49)^2)