What is the derivative of m(r)=(r^3+3r+2)/(r^2-1) ?

The derivative of m(r)=(r^3+3r+2)/(r^2-1) is (r^4-6r^2-4r-3)/((r^2-1)^2)

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

The first derivative of m(r)=(r^3+3r+2)/(r^2-1) is (r^4-6r^2-4r-3)/((r^2-1)^2)

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derivative of

m (r) = \frac{r ^{3} + 3 r + 2}{r ^{2} - 1}

\frac{r ^{4} - 6 r ^{2} - 4 r - 3}{( r ^{2} - 1 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d r} (r^{3} + 3 r + 2) = 3 r^{2} + 3

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

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

Expand

(3 r^{2} + 3) (r^{2} - 1) - 2 r (r^{3} + 3 r + 2) : r^{4} - 6 r^{2} - 4 r - 3

= \frac{r ^{4} - 6 r ^{2} - 4 r - 3}{( r ^{2} - 1 ) ^{2}}

\frac{d}{d x} (5 x - 1 - 2 x)

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

f (x) = xx

f (x) = x - 1

\frac{\partial}{\partial x} (sin (\frac{2 y}{x}))

What is the derivative of m(r)=(r^3+3r+2)/(r^2-1) ?
The derivative of m(r)=(r^3+3r+2)/(r^2-1) is (r^4-6r^2-4r-3)/((r^2-1)^2)
What is the first derivative of m(r)=(r^3+3r+2)/(r^2-1) ?
The first derivative of m(r)=(r^3+3r+2)/(r^2-1) is (r^4-6r^2-4r-3)/((r^2-1)^2)