What is the derivative of 0.125xsqrt(64-x^2) ?

The derivative of 0.125xsqrt(64-x^2) is (0.125(-2x^2+64))/(sqrt(64-x^2))

What is the first derivative of 0.125xsqrt(64-x^2) ?

The first derivative of 0.125xsqrt(64-x^2) is (0.125(-2x^2+64))/(sqrt(64-x^2))

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derivative of

0.125 x 64 - x^{2}

\frac{0.125 ( - 2 x ^{2} + 64 )}{64 - x ^{2}}

Solution steps

\frac{d}{d x} (0.125 x 64 - x^{2})

Take the constant out:

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

= 0.125 \frac{d}{d x} (x 64 - x^{2})

Apply the Product Rule:

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

f = x, g = 64 - x^{2}

= 0.125 (\frac{d x}{d x} 64 - x^{2} + \frac{d}{d x} (64 - x^{2}) x)

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

\frac{d}{d x} (64 - x^{2}) = - \frac{x}{64 - x ^{2}}

= 0.125 (1 \cdot 64 - x^{2} + (- \frac{x}{64 - x ^{2}}) x)

Simplify

0.125 (1 \cdot 64 - x^{2} + (- \frac{x}{64 - x ^{2}}) x) : \frac{0.125 ( - 2 x ^{2} + 64 )}{64 - x ^{2}}

= \frac{0.125 ( - 2 x ^{2} + 64 )}{64 - x ^{2}}

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

\frac{d}{d x} (a cos (b x))

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

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

t \frac{d y}{d t} + 5 y = t

What is the derivative of 0.125xsqrt(64-x^2) ?
The derivative of 0.125xsqrt(64-x^2) is (0.125(-2x^2+64))/(sqrt(64-x^2))
What is the first derivative of 0.125xsqrt(64-x^2) ?
The first derivative of 0.125xsqrt(64-x^2) is (0.125(-2x^2+64))/(sqrt(64-x^2))