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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derivative of

\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}}

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

\int_{- 1}^{8} 2 - 3 x d x

x \to - 1 + lim (\frac{1}{( x + 1 ) ^{2}})

\int \frac{4 x}{4 x + 1} d x

9 cos^{2} (t)

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)