What is the derivative of (9x/(64+x^2+y^2)) ?

The derivative of (9x/(64+x^2+y^2)) is (9(-x^2-2xy\frac{dy)/(dx)+y^2+64)}{(64+x^2+y^2)^2}

What is the first derivative of (9x/(64+x^2+y^2)) ?

The first derivative of (9x/(64+x^2+y^2)) is (9(-x^2-2xy\frac{dy)/(dx)+y^2+64)}{(64+x^2+y^2)^2}

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\frac{d}{d x} (\frac{9 x}{64 + x ^{2} + y ^{2}})

\frac{9 ( - x ^{2} - 2 x y \frac{d y}{d x} + y ^{2} + 64 )}{( 64 + x ^{2} + y ^{2} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{9 x}{64 + x ^{2} + y ^{2}})

Take the constant out:

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

= 9 \frac{d}{d x} (\frac{x}{64 + x ^{2} + y ^{2}})

Apply the Quotient Rule:

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

= 9 \cdot \frac{\frac{d x}{d x} ( 64 + x ^{2} + y ^{2} ) - \frac{d}{d x} ( 64 + x ^{2} + y ^{2} ) x}{( 64 + x ^{2} + y ^{2} ) ^{2}}

\frac{d x}{d x} = 1

\frac{d}{d x} (64 + x^{2} + y^{2}) = 2 x + 2 y \frac{d y}{d x}

= 9 \cdot \frac{1 \cdot ( 64 + x ^{2} + y ^{2} ) - ( 2 x + 2 y \frac{d y}{d x} ) x}{( 64 + x ^{2} + y ^{2} ) ^{2}}

Simplify

9 \cdot \frac{1 \cdot ( 64 + x ^{2} + y ^{2} ) - ( 2 x + 2 y \frac{d y}{d x} ) x}{( 64 + x ^{2} + y ^{2} ) ^{2}} : \frac{9 ( - x ^{2} - 2 x y \frac{d y}{d x} + y ^{2} + 64 )}{( 64 + x ^{2} + y ^{2} ) ^{2}}

= \frac{9 ( - x ^{2} - 2 x y \frac{d y}{d x} + y ^{2} + 64 )}{( 64 + x ^{2} + y ^{2} ) ^{2}}

\frac{d}{d x} (- ln ∣ cos (x) ∣)

f (x) = (- 3 x^{2} - 3)^{4}

t \to 0 lim (cos (3 t))

\frac{\partial}{\partial z} (\frac{( x + y )}{( y + z )})

\int \frac{e ^{θ}}{a + b e ^{θ}} d θ

What is the derivative of (9x/(64+x^2+y^2)) ?
The derivative of (9x/(64+x^2+y^2)) is (9(-x^2-2xy\frac{dy)/(dx)+y^2+64)}{(64+x^2+y^2)^2}
What is the first derivative of (9x/(64+x^2+y^2)) ?
The first derivative of (9x/(64+x^2+y^2)) is (9(-x^2-2xy\frac{dy)/(dx)+y^2+64)}{(64+x^2+y^2)^2}