What is the derivative of y=(3x^4+7)/(x^2) ?

The derivative of y=(3x^4+7)/(x^2) is (2(3x^4-7))/(x^3)

What is the first derivative of y=(3x^4+7)/(x^2) ?

The first derivative of y=(3x^4+7)/(x^2) is (2(3x^4-7))/(x^3)

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derivative of

y = \frac{3 x ^{4} + 7}{x ^{2}}

\frac{2 ( 3 x ^{4} - 7 )}{x ^{3}}

Solution steps

\frac{d}{d x} (\frac{3 x ^{4} + 7}{x ^{2}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 3 x ^{4} + 7 ) x ^{2} - \frac{d}{d x} ( x ^{2} ) ( 3 x ^{4} + 7 )}{( x ^{2} ) ^{2}}

\frac{d}{d x} (3 x^{4} + 7) = 12 x^{3}

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

= \frac{12 x ^{3} x ^{2} - 2 x ( 3 x ^{4} + 7 )}{( x ^{2} ) ^{2}}

Simplify

\frac{12 x ^{3} x ^{2} - 2 x ( 3 x ^{4} + 7 )}{( x ^{2} ) ^{2}} : \frac{2 ( 3 x ^{4} - 7 )}{x ^{3}}

= \frac{2 ( 3 x ^{4} - 7 )}{x ^{3}}

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

x^{2} d y + y d x = d x

\int x e^{6 x^{2}} d x

\int \frac{1}{sin ( x ) + tan ( x )} d x

\frac{d}{d x} (i sin (x))

What is the derivative of y=(3x^4+7)/(x^2) ?
The derivative of y=(3x^4+7)/(x^2) is (2(3x^4-7))/(x^3)
What is the first derivative of y=(3x^4+7)/(x^2) ?
The first derivative of y=(3x^4+7)/(x^2) is (2(3x^4-7))/(x^3)