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

The derivative of y=((x-3)^2-4)/(x^2+2) is (6x^2-6x-12)/((x^2+2)^2)

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

The first derivative of y=((x-3)^2-4)/(x^2+2) is (6x^2-6x-12)/((x^2+2)^2)

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{2 ( x - 3 ) ( x ^{2} + 2 ) - 2 x ( ( x - 3 ) ^{2} - 4 )}{( x ^{2} + 2 ) ^{2}} : \frac{6 x ^{2} - 6 x - 12}{( x ^{2} + 2 ) ^{2}}

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

\frac{y ^{2}}{x}

\frac{x ^{2} - 5}{1 - 3 x ^{2}}

\frac{l ^{3}}{4} k \frac{1}{4}

3 x + y

\frac{2 x}{( x ^{2} + y ^{2} )}

What is the derivative of y=((x-3)^2-4)/(x^2+2) ?
The derivative of y=((x-3)^2-4)/(x^2+2) is (6x^2-6x-12)/((x^2+2)^2)
What is the first derivative of y=((x-3)^2-4)/(x^2+2) ?
The first derivative of y=((x-3)^2-4)/(x^2+2) is (6x^2-6x-12)/((x^2+2)^2)