What is the derivative of y=4x^3sqrt(1-x^2) ?

The derivative of y=4x^3sqrt(1-x^2) is (4(-4x^4+3x^2))/(sqrt(1-x^2))

What is the first derivative of y=4x^3sqrt(1-x^2) ?

The first derivative of y=4x^3sqrt(1-x^2) is (4(-4x^4+3x^2))/(sqrt(1-x^2))

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derivative of

y = 4 x^{3} 1 - x^{2}

\frac{4 ( - 4 x ^{4} + 3 x ^{2} )}{1 - x ^{2}}

Solution steps

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

Take the constant out:

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

= 4 \frac{d}{d x} (x^{3} 1 - x^{2})

Apply the Product Rule:

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

f = x^{3}, g = 1 - x^{2}

= 4 (\frac{d}{d x} (x^{3}) 1 - x^{2} + \frac{d}{d x} (1 - x^{2}) x^{3})

\frac{d}{d x} (x^{3}) = 3 x^{2}

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

= 4 (3 x^{2} 1 - x^{2} + (- \frac{x}{1 - x ^{2}}) x^{3})

Simplify

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

= \frac{4 ( - 4 x ^{4} + 3 x ^{2} )}{1 - x ^{2}}

\int \frac{x + 4}{x - 4} d x

\frac{d}{d t} (6 - 3 t^{2})

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

\frac{\partial}{\partial y} (z - x sin (y) - 4)

x \to - 1 lim (1 - 2 x)

What is the derivative of y=4x^3sqrt(1-x^2) ?
The derivative of y=4x^3sqrt(1-x^2) is (4(-4x^4+3x^2))/(sqrt(1-x^2))
What is the first derivative of y=4x^3sqrt(1-x^2) ?
The first derivative of y=4x^3sqrt(1-x^2) is (4(-4x^4+3x^2))/(sqrt(1-x^2))