What is the derivative of (x^2+2x-6/((x+1)^2)) ?

The derivative of (x^2+2x-6/((x+1)^2)) is (14)/((x+1)^3)

What is the first derivative of (x^2+2x-6/((x+1)^2)) ?

The first derivative of (x^2+2x-6/((x+1)^2)) is (14)/((x+1)^3)

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\frac{d}{d x} (\frac{x ^{2} + 2 x - 6}{( x + 1 ) ^{2}})

\frac{14}{( x + 1 ) ^{3}}

Solution steps

\frac{d}{d x} (\frac{x ^{2} + 2 x - 6}{( x + 1 ) ^{2}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{2} + 2 x - 6 ) ( x + 1 ) ^{2} - \frac{d}{d x} ( ( x + 1 ) ^{2} ) ( x ^{2} + 2 x - 6 )}{( ( x + 1 ) ^{2} ) ^{2}}

\frac{d}{d x} (x^{2} + 2 x - 6) = 2 x + 2

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

= \frac{( 2 x + 2 ) ( x + 1 ) ^{2} - 2 ( x + 1 ) ( x ^{2} + 2 x - 6 )}{( ( x + 1 ) ^{2} ) ^{2}}

Simplify

\frac{( 2 x + 2 ) ( x + 1 ) ^{2} - 2 ( x + 1 ) ( x ^{2} + 2 x - 6 )}{( ( x + 1 ) ^{2} ) ^{2}} : \frac{14}{( x + 1 ) ^{3}}

= \frac{14}{( x + 1 ) ^{3}}

(x^{2} + 1) y^{^{''}} - 2 x y^{^{'}} + 2 y = 0

f (x) = 6 e^{x} + x, (0, 6)

\frac{d}{d x} (cos^{4} (2 x^{3} - 5))

\int \frac{5 - 2 x ^{2}}{y ^{2}} d y

\int sin (10 x) d x

What is the derivative of (x^2+2x-6/((x+1)^2)) ?
The derivative of (x^2+2x-6/((x+1)^2)) is (14)/((x+1)^3)
What is the first derivative of (x^2+2x-6/((x+1)^2)) ?
The first derivative of (x^2+2x-6/((x+1)^2)) is (14)/((x+1)^3)