What is the derivative of f(x)=(x^3-8)/(x^2+3) ?

The derivative of f(x)=(x^3-8)/(x^2+3) is (x^4+9x^2+16x)/((x^2+3)^2)

What is the first derivative of f(x)=(x^3-8)/(x^2+3) ?

The first derivative of f(x)=(x^3-8)/(x^2+3) is (x^4+9x^2+16x)/((x^2+3)^2)

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derivative of

f (x) = \frac{x ^{3} - 8}{x ^{2} + 3}

\frac{x ^{4} + 9 x ^{2} + 16 x}{( x ^{2} + 3 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{3} - 8}{x ^{2} + 3})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{3} - 8 ) ( x ^{2} + 3 ) - \frac{d}{d x} ( x ^{2} + 3 ) ( x ^{3} - 8 )}{( x ^{2} + 3 ) ^{2}}

\frac{d}{d x} (x^{3} - 8) = 3 x^{2}

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

= \frac{3 x ^{2} ( x ^{2} + 3 ) - 2 x ( x ^{3} - 8 )}{( x ^{2} + 3 ) ^{2}}

Expand

3 x^{2} (x^{2} + 3) - 2 x (x^{3} - 8) : x^{4} + 9 x^{2} + 16 x

= \frac{x ^{4} + 9 x ^{2} + 16 x}{( x ^{2} + 3 ) ^{2}}

d er i v a t i v e f (t) = e^{4 t s i n (2 t)}

\int \frac{sin ( 2 x )}{x ^{2}} d x

\int x^{3} ln (6 x) d x

\int \frac{y - 1}{y + 4} d y

\int 8 sec^{3} (6 x) d x

What is the derivative of f(x)=(x^3-8)/(x^2+3) ?
The derivative of f(x)=(x^3-8)/(x^2+3) is (x^4+9x^2+16x)/((x^2+3)^2)
What is the first derivative of f(x)=(x^3-8)/(x^2+3) ?
The first derivative of f(x)=(x^3-8)/(x^2+3) is (x^4+9x^2+16x)/((x^2+3)^2)