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

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

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

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

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derivative of

\frac{x ^{2} - 1}{x + 3}

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

= \frac{x ^{2} + 6 x + 1}{( x + 3 ) ^{2}}

e^{2 - t}

(t - 4)^{2}

\frac{\partial}{\partial y} (x^{3})

(x^{4} + y^{4}) d x - x y^{3} d y = 0

\frac{d y}{d x} - tan (x) - \frac{1}{x} = 0

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