What is the derivative of (x^2+14/(x^8)) ?

The derivative of (x^2+14/(x^8)) is (2(-3x^2-56))/(x^9)

What is the first derivative of (x^2+14/(x^8)) ?

The first derivative of (x^2+14/(x^8)) is (2(-3x^2-56))/(x^9)

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\frac{d}{d x} (\frac{x ^{2} + 14}{x ^{8}})

\frac{2 ( - 3 x ^{2} - 56 )}{x ^{9}}

Solution steps

\frac{d}{d x} (\frac{x ^{2} + 14}{x ^{8}})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( x ^{2} + 14 ) x ^{8} - \frac{d}{d x} ( x ^{8} ) ( x ^{2} + 14 )}{( x ^{8} ) ^{2}}

\frac{d}{d x} (x^{2} + 14) = 2 x

\frac{d}{d x} (x^{8}) = 8 x^{7}

= \frac{2 x x ^{8} - 8 x ^{7} ( x ^{2} + 14 )}{( x ^{8} ) ^{2}}

Simplify

\frac{2 x x ^{8} - 8 x ^{7} ( x ^{2} + 14 )}{( x ^{8} ) ^{2}} : \frac{2 ( - 3 x ^{2} - 56 )}{x ^{9}}

= \frac{2 ( - 3 x ^{2} - 56 )}{x ^{9}}

y^{^{''}} + 4 y^{^{'}} + 6.25 y = 3.125 (x + 1)^{2}

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

\frac{d}{d x} (ln^{2} (1 - 2 x))

f (x) = \frac{x}{ln ( x )}

y^{^{''}} + 6 y^{^{'}} + 19 y = 0

What is the derivative of (x^2+14/(x^8)) ?
The derivative of (x^2+14/(x^8)) is (2(-3x^2-56))/(x^9)
What is the first derivative of (x^2+14/(x^8)) ?
The first derivative of (x^2+14/(x^8)) is (2(-3x^2-56))/(x^9)