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

The derivative of f(x)=(x+1)(x+1-1/(x+2)) is (2x^3+8x^2+8x-1)/((x+2)^2)+2

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

The first derivative of f(x)=(x+1)(x+1-1/(x+2)) is (2x^3+8x^2+8x-1)/((x+2)^2)+2

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derivative of

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

\frac{2 x ^{3} + 8 x ^{2} + 8 x - 1}{( x + 2 ) ^{2}} + 2

Solution steps

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x + 1, g = x + 1 - \frac{1}{x + 2}

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

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

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

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

Simplify

1 \cdot (x + 1 - \frac{1}{x + 2}) + (\frac{1}{( x + 2 ) ^{2}} + 1) (x + 1) : \frac{2 x ^{3} + 8 x ^{2} + 8 x - 1}{( x + 2 ) ^{2}} + 2

= \frac{2 x ^{3} + 8 x ^{2} + 8 x - 1}{( x + 2 ) ^{2}} + 2

\frac{s - 1}{s ( s + 2.5 )}

\frac{\partial}{\partial x} (3 x z e^{4 x y})

x \to - 3 lim (3 x^{2} - 4 x + 1)

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

x \to - 1 lim ((x + 2 x^{3})^{5})

What is the derivative of f(x)=(x+1)(x+1-1/(x+2)) ?
The derivative of f(x)=(x+1)(x+1-1/(x+2)) is (2x^3+8x^2+8x-1)/((x+2)^2)+2
What is the first derivative of f(x)=(x+1)(x+1-1/(x+2)) ?
The first derivative of f(x)=(x+1)(x+1-1/(x+2)) is (2x^3+8x^2+8x-1)/((x+2)^2)+2