What is the derivative of (4x^3-5/(x^2-3)) ?

The derivative of (4x^3-5/(x^2-3)) is (4x^4-36x^2+10x)/((x^2-3)^2)

What is the first derivative of (4x^3-5/(x^2-3)) ?

The first derivative of (4x^3-5/(x^2-3)) is (4x^4-36x^2+10x)/((x^2-3)^2)

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\frac{d}{d x} (\frac{4 x ^{3} - 5}{x ^{2} - 3})

\frac{4 x ^{4} - 36 x ^{2} + 10 x}{( x ^{2} - 3 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{4 x ^{3} - 5}{x ^{2} - 3})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 4 x ^{3} - 5 ) ( x ^{2} - 3 ) - \frac{d}{d x} ( x ^{2} - 3 ) ( 4 x ^{3} - 5 )}{( x ^{2} - 3 ) ^{2}}

\frac{d}{d x} (4 x^{3} - 5) = 12 x^{2}

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

= \frac{12 x ^{2} ( x ^{2} - 3 ) - 2 x ( 4 x ^{3} - 5 )}{( x ^{2} - 3 ) ^{2}}

Expand

12 x^{2} (x^{2} - 3) - 2 x (4 x^{3} - 5) : 4 x^{4} - 36 x^{2} + 10 x

= \frac{4 x ^{4} - 36 x ^{2} + 10 x}{( x ^{2} - 3 ) ^{2}}

x \to 0 - lim (\frac{6}{x})

\frac{d}{d x} (4 x x)

\int \frac{6}{6 x + 1} d x

f (x) = sec (x), at x = - \frac{7 π}{4}

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

What is the derivative of (4x^3-5/(x^2-3)) ?
The derivative of (4x^3-5/(x^2-3)) is (4x^4-36x^2+10x)/((x^2-3)^2)
What is the first derivative of (4x^3-5/(x^2-3)) ?
The first derivative of (4x^3-5/(x^2-3)) is (4x^4-36x^2+10x)/((x^2-3)^2)