Question

find derivative of x+1/x whole cube

Answer 1

x³+1/x³+3(x+1/x)
hope it will help you

Answer 2

Answer:

Derivative of $(x+\frac{1}{x})^3$ is $3x^2-\frac{3}{x^4}+3-\frac{3}{x^2}$

Step-by-step explanation:

Given: $(x+\frac{1}{x})^3$

To find: Derivative w.r.t x of given function

First we simply the expression using identity,

(a+b)³ = a³ + b³ + 3a²b + 3ab²

$(x+\frac{1}{x})^3=x^3+(\frac{1}{x})^3+3\times x^2\times\frac{1}{x}+3\times x\times(\frac{1}{x})^2$

$(x+\frac{1}{x})^3=x^3+\frac{1}{x^3}+3x+\frac{3}{x}$

Now,

$\frac{\mathrm{d}(x^3+\frac{1}{x^3}+3x+\frac{3}{x})}{\mathrm{d} x}$

$=\frac{\mathrm{d}(x^3)}{\mathrm{d} x}+\frac{\mathrm{d} (\frac{1}{x^3})}{\mathrm{d} x}+\frac{\mathrm{d} (3x)}{\mathrm{d} x}+\frac{\mathrm{d} (\frac{3}{x})}{\mathrm{d} x}$

$=3x^2-3x^{-4}+3-3x^{-2}$

$=3x^2-3\frac{1}{x^4}+3-3\frac{1}{x^2}$

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

Therefore, Derivative of $(x+\frac{1}{x})^3$ is $3x^2-\frac{3}{x^4}+3-\frac{3}{x^2}$