What is the derivative of (x^4+9/(x^2)) ?

The derivative of (x^4+9/(x^2)) is (2(x^4-9))/(x^3)

What is the first derivative of (x^4+9/(x^2)) ?

The first derivative of (x^4+9/(x^2)) is (2(x^4-9))/(x^3)

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\frac{d}{d x} (\frac{x ^{4} + 9}{x ^{2}})

\frac{2 ( x ^{4} - 9 )}{x ^{3}}

Solution steps

\frac{d}{d x} (\frac{x ^{4} + 9}{x ^{2}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{4} + 9 ) x ^{2} - \frac{d}{d x} ( x ^{2} ) ( x ^{4} + 9 )}{( x ^{2} ) ^{2}}

\frac{d}{d x} (x^{4} + 9) = 4 x^{3}

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

= \frac{4 x ^{3} x ^{2} - 2 x ( x ^{4} + 9 )}{( x ^{2} ) ^{2}}

Simplify

\frac{4 x ^{3} x ^{2} - 2 x ( x ^{4} + 9 )}{( x ^{2} ) ^{2}} : \frac{2 ( x ^{4} - 9 )}{x ^{3}}

= \frac{2 ( x ^{4} - 9 )}{x ^{3}}

sin (e^{x})

\frac{d}{d x} (3 \frac{8}{x})

\int_{0}^{3} \frac{1}{x ^{2} + 9} d x

x y^{^{'}} + 2 y = 3

\frac{d}{d x} (\frac{x}{x ^{2} + 6})

What is the derivative of (x^4+9/(x^2)) ?
The derivative of (x^4+9/(x^2)) is (2(x^4-9))/(x^3)
What is the first derivative of (x^4+9/(x^2)) ?
The first derivative of (x^4+9/(x^2)) is (2(x^4-9))/(x^3)