What is the derivative of (9sqrt(x+8))x^2 ?

The derivative of (9sqrt(x+8))x^2 is (9(5x^2+32x))/(2sqrt(x+8))

What is the first derivative of (9sqrt(x+8))x^2 ?

The first derivative of (9sqrt(x+8))x^2 is (9(5x^2+32x))/(2sqrt(x+8))

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derivative of

(9 x + 8) x^{2}

\frac{9 ( 5 x ^{2} + 32 x )}{2 x + 8}

Solution steps

\frac{d}{d x} (9 x + 8 x^{2})

Take the constant out:

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

= 9 \frac{d}{d x} (x + 8 x^{2})

Apply the Product Rule:

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

f = x + 8, g = x^{2}

= 9 (\frac{d}{d x} (x + 8) x^{2} + \frac{d}{d x} (x^{2}) x + 8)

\frac{d}{d x} (x + 8) = \frac{1}{2 x + 8}

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

= 9 (\frac{1}{2 x + 8} x^{2} + 2 x x + 8)

Simplify

9 (\frac{1}{2 x + 8} x^{2} + 2 x x + 8) : \frac{9 ( 5 x ^{2} + 32 x )}{2 x + 8}

= \frac{9 ( 5 x ^{2} + 32 x )}{2 x + 8}

\int_{0}^{\infty} \frac{y e ^{\frac{- y}{a}}}{a} d y

y^{^{''}} - 2 y^{^{'}} + y = 2 e^{x}

f (x) = sin (sin (x)), at x = 2 π

x \to - 1 + lim (arcsin (x))

f (x) = ln (\frac{x + 3}{x - 3})

What is the derivative of (9sqrt(x+8))x^2 ?
The derivative of (9sqrt(x+8))x^2 is (9(5x^2+32x))/(2sqrt(x+8))
What is the first derivative of (9sqrt(x+8))x^2 ?
The first derivative of (9sqrt(x+8))x^2 is (9(5x^2+32x))/(2sqrt(x+8))