What is the derivative of x/(sqrt(36+x^2)) ?

The derivative of x/(sqrt(36+x^2)) is (36)/((36+x^2)sqrt(36+x^2))

What is the first derivative of x/(sqrt(36+x^2)) ?

The first derivative of x/(sqrt(36+x^2)) is (36)/((36+x^2)sqrt(36+x^2))

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\frac{d}{d x} (\frac{x}{36 + x ^{2}})

\frac{36}{( 36 + x ^{2} ) 36 + x ^{2}}

Solution steps

\frac{d}{d x} (\frac{x}{36 + x ^{2}})

Apply the Quotient Rule:

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

= \frac{\frac{d x}{d x} 36 + x ^{2} - \frac{d}{d x} ( 36 + x ^{2} ) x}{( 36 + x ^{2} ) ^{2}}

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

\frac{d}{d x} (36 + x^{2}) = \frac{x}{36 + x ^{2}}

= \frac{1 \cdot 36 + x ^{2} - \frac{x}{36 + x ^{2}} x}{( 36 + x ^{2} ) ^{2}}

Simplify

\frac{1 \cdot 36 + x ^{2} - \frac{x}{36 + x ^{2}} x}{( 36 + x ^{2} ) ^{2}} : \frac{36}{( 36 + x ^{2} ) 36 + x ^{2}}

= \frac{36}{( 36 + x ^{2} ) 36 + x ^{2}}

\frac{\partial}{\partial x} (2 x e^{x^{2}} + y^{2})

y^{^{''}} + 6 y^{^{'}} + 18 y = 0

y \to 0 lim (cot (5 y))

\frac{\partial}{\partial x} (e^{6 y} cos (- 6 x))

\frac{d y}{d x} = (\frac{1 + 2 y ^{2}}{y sin ( x )})

What is the derivative of x/(sqrt(36+x^2)) ?
The derivative of x/(sqrt(36+x^2)) is (36)/((36+x^2)sqrt(36+x^2))
What is the first derivative of x/(sqrt(36+x^2)) ?
The first derivative of x/(sqrt(36+x^2)) is (36)/((36+x^2)sqrt(36+x^2))