I've differentiated the $\frac{1-x}{1+x}$ which is $\frac{-2}{(1+x)^2}$ I don't know how to proceed to it. Hope someone can show it. Thanks in advance.
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2$\begingroup$ We have $y = \sqrt{1 - x \over 1 + x}$, so you know that $y^2 = {1 - x \over 1 + x}.$ Then you simply take the derivative of both sides to give you $2yy' = \frac{-2}{(1+x)^2}$ (Which we receive this by implicit differentiation). $\endgroup$– Decaf-MathCommented Sep 26, 2016 at 8:49
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2 Answers
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Let $$y=\sqrt{\frac{1-x}{1+x}}=\sqrt{f}$$ Then $$\frac{dy}{dx}=\frac{1}{2}\frac{1}{\sqrt{f}}\frac{df}{dx}$$ So $$\frac{dy}{dx}=\frac{1}{2}\frac{1}{\sqrt{\frac{1-x}{1+x}}}\frac{-2}{(1+x)^2}=\frac{-1}{(1+x)^{\frac{3}{2}}\sqrt{1-x}}=\frac{-1}{(1+x)\sqrt{1-x^2}}$$ Since if two curves are perpendicular, the product of their gradients is -1, so the gradient of the normal to this question is $(1+x)\sqrt{1-x^2}$. Since $$\frac{dy}{dx}(1+x)\sqrt{1-x^2}=-1$$
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Hint. One may apply the chain rule $$ \left(\sqrt{f}\right)'=f' \cdot \frac{1}{2\sqrt{f}} $$ to get $$ \left(\sqrt{\frac{1-x}{1+x}}\right)'=\frac{-2}{(1+x)^2} \cdot \frac{1}{2\sqrt{\frac{1-x}{1+x}}}=\frac{-1}{(1+x)^{3/2}\sqrt{1-x}},\qquad x<1. $$
answered Sep 26, 2016 at 8:47
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$\begingroup$ This answer is perfectly fine, but it clearly doesn't demonstrate the derivative in the fashion that the OP was looking for--why the book gave the hint of taking the derivative of the squared quantity. $\endgroup$ Commented Sep 26, 2016 at 8:51
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$\begingroup$ @pyrazolam Sorry, apparently I did not read correctly the OP question. Hoping someone will give an appropriate answer. I leave my answer above which is correct. Thanks for your feedback. $\endgroup$ Commented Sep 26, 2016 at 8:56