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Question

Mathematics Question on limits and derivatives

If h(x)=2+x22−x2h\left(x\right)=\frac{2+x^{2}}{2-x^{2}}, h′(1)=h'\left(1\right)=

A

22

B

44

C

66

D

88

Answer

88

Explanation

Solution

We have h(x)=2+x22−x2h\left(x\right)=\frac{2+x^{2}}{2-x^{2}} ∴h′(x)=(2−x2)(2x)−(2+x2)(−2x)(2−x2)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}} =2x(2−x2+2+x2)(2−x2)2=\frac{2x\left(2-x^{2}+2+x^{2}\right)}{\left(2-x^{2}\right)^{2}} =8x(2−x2)2=\frac{8x}{\left(2-x^{2}\right)^{2}} ∴h′(1)=81=8\therefore h'\left(1\right)=\frac{8}{1}=8.