If |x|<1 and y=1+x+x^2+x^3+ , then write the value of dy dx .

Question

If $|\text{x}|<1$ and $\text{y}=1+\text{x}+\text{x}^2+\text{x}^3+\dots,$ then write the value of $\frac{\text{dy}}{\text{dx}}.$

✓

Answer

We have,$\text{y}=1+\text{x}+\text{x}^2+\text{x}^3+\dots$
$=(1-\text{x})^{-1}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(1-\text{x})^{-1}$
$={-1}(1-\text{x})^{-2}\times(-1)$
$=(1-\text{x})^{-2}$
$=\frac{1}{(1-\text{x})^{2}}$

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