What is the derivative of (6x^2+1/(x^2+8)) ?

The derivative of (6x^2+1/(x^2+8)) is (94x)/((x^2+8)^2)

What is the first derivative of (6x^2+1/(x^2+8)) ?

The first derivative of (6x^2+1/(x^2+8)) is (94x)/((x^2+8)^2)

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\frac{d}{d x} (\frac{6 x ^{2} + 1}{x ^{2} + 8})

\frac{94 x}{( x ^{2} + 8 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{6 x ^{2} + 1}{x ^{2} + 8})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 6 x ^{2} + 1 ) ( x ^{2} + 8 ) - \frac{d}{d x} ( x ^{2} + 8 ) ( 6 x ^{2} + 1 )}{( x ^{2} + 8 ) ^{2}}

\frac{d}{d x} (6 x^{2} + 1) = 12 x

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

= \frac{12 x ( x ^{2} + 8 ) - 2 x ( 6 x ^{2} + 1 )}{( x ^{2} + 8 ) ^{2}}

Expand

12 x (x^{2} + 8) - 2 x (6 x^{2} + 1) : 94 x

= \frac{94 x}{( x ^{2} + 8 ) ^{2}}

y = cos ((π x)^{8})

\int_{- 1}^{2} x^{2} (x^{3} + 1)^{2} d x

\frac{d y}{d x} = 100 x (y + 2)

y = 3 e^{x} cos (x), (0, 3)

4 y^{^{''}} + 12 y^{^{'}} + 9 y = 0

What is the derivative of (6x^2+1/(x^2+8)) ?
The derivative of (6x^2+1/(x^2+8)) is (94x)/((x^2+8)^2)
What is the first derivative of (6x^2+1/(x^2+8)) ?
The first derivative of (6x^2+1/(x^2+8)) is (94x)/((x^2+8)^2)