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

The derivative of (x^4-2/(x^2-2)) is (2x^5-8x^3+4x)/((x^2-2)^2)

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

The first derivative of (x^4-2/(x^2-2)) is (2x^5-8x^3+4x)/((x^2-2)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\frac{2 x ^{5} - 8 x ^{3} + 4 x}{( x ^{2} - 2 ) ^{2}}

Solution steps

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

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

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

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

Expand

4 x^{3} (x^{2} - 2) - 2 x (x^{4} - 2) : 2 x^{5} - 8 x^{3} + 4 x

= \frac{2 x ^{5} - 8 x ^{3} + 4 x}{( x ^{2} - 2 ) ^{2}}

\frac{\partial}{\partial x} (6 x^{2} y + 6 x y^{3})

x \to - 5 lim (g (x))

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

\frac{s + 1}{( s ^{2} - 2 s + 5 )}

\frac{\partial}{\partial x} (x yz + x y^{5} + y z^{5} + z x^{5})

What is the derivative of (x^4-2/(x^2-2)) ?
The derivative of (x^4-2/(x^2-2)) is (2x^5-8x^3+4x)/((x^2-2)^2)
What is the first derivative of (x^4-2/(x^2-2)) ?
The first derivative of (x^4-2/(x^2-2)) is (2x^5-8x^3+4x)/((x^2-2)^2)