What is the derivative of f(x)= x/(sqrt(1-x^2)) ?

The derivative of f(x)= x/(sqrt(1-x^2)) is 1/((1-x^2)sqrt(1-x^2))

What is the first derivative of f(x)= x/(sqrt(1-x^2)) ?

The first derivative of f(x)= x/(sqrt(1-x^2)) is 1/((1-x^2)sqrt(1-x^2))

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derivative of

f (x) = \frac{x}{1 - x ^{2}}

\frac{1}{( 1 - x ^{2} ) 1 - x ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

f (x) = \frac{2500 x}{ln ( 8 x + 8 )}

x y^{^{'}} + (x + \frac{1}{3}) y = e^{- 2 x} y^{- 2}

f (x) = x^{2} - x + 4, at x = 1

f (x) = x^{2} - 4, (4, 12)

What is the derivative of f(x)= x/(sqrt(1-x^2)) ?
The derivative of f(x)= x/(sqrt(1-x^2)) is 1/((1-x^2)sqrt(1-x^2))
What is the first derivative of f(x)= x/(sqrt(1-x^2)) ?
The first derivative of f(x)= x/(sqrt(1-x^2)) is 1/((1-x^2)sqrt(1-x^2))