What is the derivative of f(x)=xsqrt(x-x^2) ?

The derivative of f(x)=xsqrt(x-x^2) is (-4x^2+3x)/(2sqrt(x-x^2))

What is the first derivative of f(x)=xsqrt(x-x^2) ?

The first derivative of f(x)=xsqrt(x-x^2) is (-4x^2+3x)/(2sqrt(x-x^2))

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derivative of

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

\frac{- 4 x ^{2} + 3 x}{2 x - x ^{2}}

Solution steps

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = x - x^{2}

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

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

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

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

Simplify

1 \cdot x - x^{2} + \frac{1 - 2 x}{2 x - x ^{2}} x : \frac{- 4 x ^{2} + 3 x}{2 x - x ^{2}}

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

e^{x}, ln (12)

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

x \to 0 lim (\frac{9}{x})

\frac{d y}{d x} = - \frac{1 + x + y}{1 + x}

y^{^{''}} - 2 y^{^{'}} + 5 y = 0, y (\frac{π}{2}) = 0

What is the derivative of f(x)=xsqrt(x-x^2) ?
The derivative of f(x)=xsqrt(x-x^2) is (-4x^2+3x)/(2sqrt(x-x^2))
What is the first derivative of f(x)=xsqrt(x-x^2) ?
The first derivative of f(x)=xsqrt(x-x^2) is (-4x^2+3x)/(2sqrt(x-x^2))