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Mathematics 21 Online
OpenStudy (anonymous):

Why is the derivative of (x^2/8) equal to (1/4)x?

Parth (parthkohli):

\[ \dfrac{x^2}{8} = x^2 \times \dfrac{1}{8}\]Product Rule.

Parth (parthkohli):

Or even the Power Rule.

OpenStudy (anonymous):

\[f(x) = 3x ^{4}-6x ^{3}+\frac{ x ^{2} }{ 8 }+5\] \[f'(x)=12x ^{3}-18x ^{2}+\frac{ 1 }{ 4 }x\] How did I get the 1/4? I'm having trouble figuring out how to apply the power rule to fractions.

OpenStudy (zarkon):

2/8=1/4

OpenStudy (anonymous):

Derp. Thank you!

Parth (parthkohli):

\[\dfrac{d}{dx} \left( \dfrac{1}{8} x^2 \right)= \dfrac{d}{dx} \left(\dfrac{1}{8} \times 2x\right) = \dfrac{1}{4} x\]

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