What is the derivative of y=(sqrt(x))/(9+x) ?

The derivative of y=(sqrt(x))/(9+x) is (9-x)/(2sqrt(x)(9+x)^2)

What is the first derivative of y=(sqrt(x))/(9+x) ?

The first derivative of y=(sqrt(x))/(9+x) is (9-x)/(2sqrt(x)(9+x)^2)

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derivative of

y = \frac{x}{9 + x}

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

Solution steps

\frac{d}{d x} (\frac{x}{9 + x})

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (9 + x) = 1

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

Simplify

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

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

x \to 0 lim (\frac{0}{1})

(- 14 x y^{2} + 19 y) + (- 14 x^{2} y + 19 x) y^{^{'}} = 0

\int_{1}^{3} x 4^{(x^{2})} d x

6 t

x \to 0 lim (\frac{2 tan ( 2 x )}{x})

What is the derivative of y=(sqrt(x))/(9+x) ?
The derivative of y=(sqrt(x))/(9+x) is (9-x)/(2sqrt(x)(9+x)^2)
What is the first derivative of y=(sqrt(x))/(9+x) ?
The first derivative of y=(sqrt(x))/(9+x) is (9-x)/(2sqrt(x)(9+x)^2)