Question

 यदि y = ${\displaystyle \log \left(\frac{1-x^2}{1+x^2}\right)}$ है, तो ${\displaystyle \frac{\text{dy}}{\text{dx}}}$ बराबर है।

Options

• ${\displaystyle \frac{4x^3}{1-x^4}}$

• ${\displaystyle \frac{-4x}{1-x^4}}$

• ${\displaystyle \frac{1}{4-x^4}}$

• ${\displaystyle \frac{-4x^3}{1-x^4}}$

Answer 1

सही उत्तर ${\displaystyle \underline{\frac{-4x}{1-x^4}}}$  है।

 व्याख्या:

यह देखते हुए: y = ${\displaystyle \log \left(\frac{1-x^2}{1+x^2}\right)}$  

⇒ y = log(1 – x²) – log(1 + x²)     ....${\displaystyle [\text{क्योंकि}\log \frac{x}{y}=\log x-\log y]}$

दोनों पक्षों को अलग करते हुए w.r.t. x

${\displaystyle \frac{\text{dy}}{\text{dx}}=\frac{1}{1-x^2}\cdot \frac{\text{d}}{\text{dx}}(1-x^2)-\frac{1}{1+x^2}(1+x^2)}$

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

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

= ${\displaystyle \frac{-4x}{1-x^4}}$.