What is the derivative of y=(x^2+32sqrt(x))/8 ?

The derivative of y=(x^2+32sqrt(x))/8 is 1/8 (2x+(16)/(sqrt(x)))

What is the first derivative of y=(x^2+32sqrt(x))/8 ?

The first derivative of y=(x^2+32sqrt(x))/8 is 1/8 (2x+(16)/(sqrt(x)))

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derivative of

y = \frac{x ^{2} + 32 x}{8}

\frac{1}{8} (2 x + \frac{16}{x})

Solution steps

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

Take the constant out:

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

= \frac{1}{8} \frac{d}{d x} (x^{2} + 32 x)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{1}{8} (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (32 x))

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

\frac{d}{d x} (32 x) = \frac{16}{x}

= \frac{1}{8} (2 x + \frac{16}{x})

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What is the derivative of y=(x^2+32sqrt(x))/8 ?
The derivative of y=(x^2+32sqrt(x))/8 is 1/8 (2x+(16)/(sqrt(x)))
What is the first derivative of y=(x^2+32sqrt(x))/8 ?
The first derivative of y=(x^2+32sqrt(x))/8 is 1/8 (2x+(16)/(sqrt(x)))