Question:

If $h\left(x\right)=\frac{2+x^{2}}{2-x^{2}}$, $h'\left(1\right)=$

Updated On: Jul 6, 2022
  • $2$
  • $4$
  • $6$
  • $8$
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The Correct Option is D

Solution and Explanation

We have $h\left(x\right)=\frac{2+x^{2}}{2-x^{2}}$ $\therefore h'\left(x\right)=\frac{\left(2-x^{2}\right)\left(2x\right)-\left(2+x^{2}\right)\left(-2x\right)}{\left(2-x^{2}\right)^{2}}$ $=\frac{2x\left(2-x^{2}+2+x^{2}\right)}{\left(2-x^{2}\right)^{2}}$ $=\frac{8x}{\left(2-x^{2}\right)^{2}}$ $\therefore h'\left(1\right)=\frac{8}{1}=8$.
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