What is the derivative of (8*sqrt(x))/(x^2+p) ?

The derivative of (8*sqrt(x))/(x^2+p) is (4(p-3x^2))/(sqrt(x)(x^2+p)^2)

What is the first derivative of (8*sqrt(x))/(x^2+p) ?

The first derivative of (8*sqrt(x))/(x^2+p) is (4(p-3x^2))/(sqrt(x)(x^2+p)^2)

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derivative of

\frac{8 \cdot x}{x ^{2} + p}

\frac{4 ( p - 3 x ^{2} )}{x ( x ^{2} + p ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{8 x}{x ^{2} + p})

Treat

p

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 8 \frac{d}{d x} (\frac{x}{x ^{2} + p})

Apply the Quotient Rule:

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

= 8 \cdot \frac{\frac{d}{d x} ( x ) ( x ^{2} + p ) - \frac{d}{d x} ( x ^{2} + p ) x}{( x ^{2} + p ) ^{2}}

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

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

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

Simplify

8 \cdot \frac{\frac{1}{2 x} ( x ^{2} + p ) - 2 x x}{( x ^{2} + p ) ^{2}} : \frac{4 ( p - 3 x ^{2} )}{x ( x ^{2} + p ) ^{2}}

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

\int e^{x} \cdot e^{x} d x

n = 1 \sum \infty \frac{4}{5 ^{n}}

6 x + \frac{8}{3}

\int \frac{ln ^{2} ( 5 )}{2 x} d x

\int_{0}^{\frac{π}{4}} x sin (16 x) d x

What is the derivative of (8*sqrt(x))/(x^2+p) ?
The derivative of (8*sqrt(x))/(x^2+p) is (4(p-3x^2))/(sqrt(x)(x^2+p)^2)
What is the first derivative of (8*sqrt(x))/(x^2+p) ?
The first derivative of (8*sqrt(x))/(x^2+p) is (4(p-3x^2))/(sqrt(x)(x^2+p)^2)