What is the derivative of 8(x-1/x)^4 ?

The derivative of 8(x-1/x)^4 is 32(x-1/x)^3(1+1/(x^2))

What is the first derivative of 8(x-1/x)^4 ?

The first derivative of 8(x-1/x)^4 is 32(x-1/x)^3(1+1/(x^2))

Popular >

derivative of

8 (x - \frac{1}{x})^{4}

32 (x - \frac{1}{x})^{3} (1 + \frac{1}{x ^{2}})

Solution steps

\frac{d}{d x} (8 (x - \frac{1}{x})^{4})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 8 \frac{d}{d x} ((x - \frac{1}{x})^{4})

Apply the chain rule:

4 (x - \frac{1}{x})^{3} \frac{d}{d x} (x - \frac{1}{x})

= 4 (x - \frac{1}{x})^{3} \frac{d}{d x} (x - \frac{1}{x})

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

= 8 \cdot 4 (x - \frac{1}{x})^{3} (1 + \frac{1}{x ^{2}})

Simplify

= 32 (x - \frac{1}{x})^{3} (1 + \frac{1}{x ^{2}})

\frac{\partial}{\partial y} (x e^{y})

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

\int_{0}^{1} [(x^{2} - 1) - (- x + 2)] d x

\frac{\partial}{\partial x} (2 x - y)

\int 2 x (x^{2} + 2)^{5} d x

What is the derivative of 8(x-1/x)^4 ?
The derivative of 8(x-1/x)^4 is 32(x-1/x)^3(1+1/(x^2))
What is the first derivative of 8(x-1/x)^4 ?
The first derivative of 8(x-1/x)^4 is 32(x-1/x)^3(1+1/(x^2))