What is the derivative of 2xsqrt(36-x^2) ?

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

What is the first derivative of 2xsqrt(36-x^2) ?

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

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

f = x, g = 36 - x^{2}

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

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

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

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

Simplify

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

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

x y^{^{'}} + 3 y^{^{'}} - (x + 3) sin (x) + y = 0

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

\frac{d}{d x} (1 8^{x})

x \to 0 lim (cos (19 x))

\frac{\partial}{\partial x} (ln (x^{3} y + 2 z))

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