[SOLUTION: Like Tom-Himler said: numerator is wrong because chain-rule was not followed]
I am reading, Calculus Made Easy and there is this example on page 73:
$$\omega = \frac{\sqrt{1-\theta^2}}{1+\theta}$$ (notice square-root is only in numerator). The author made the calculation easier by cancelling out, $$\omega = \sqrt\frac{1-\theta}{1+\theta}$$ (notice square-root is in both numerator and denominator) and then taking the derivative $$\frac{d\omega}{d\theta} = \frac{-1}{(1+\theta)\sqrt{1-\theta^2}}$$
But when I do the derivative without cancelling out, it does not come out to be correct.
Quotient rules states that the derivative of a function that is the ratio of two different functions is $$\frac{d}{dx}\frac{u(x)}{v(x)} = \frac{(v \frac{du}{dx}) -(u\frac{dv}{dx})}{v^2}$$
Hence $u = (1 - \theta^2), v = (1+\theta)$
$$\frac{d\omega}{d\theta}$$ evaluates to
Numerator $$\frac{(1+\theta)}{2\cdot \sqrt{1-\theta^2}} - \sqrt {1 - \theta^2}\cdot 1$$
Denominator $$(1+\theta)^2 = (1+\theta)(1+\theta)$$
Simplifying numerator: $$\frac{(1+\theta) - 2(1-\theta^2)}{2\sqrt{1 - \theta^2}}$$
Putting back numerator and denominator $$\frac{(1+\theta) -2(1+\theta)(1-\theta)}{2(1+\theta)(1+\theta)\sqrt{1-\theta^2}}\\ \frac{1-2(1-\theta)}{2(1+\theta)\sqrt{1-\theta^2}}\\ \frac{1-2+2\theta}{2(1+\theta)\sqrt{1-\theta^2}}\\ \frac{-1+2\theta}{2(1+\theta)\sqrt{1-\theta^2}}$$
This does not match the answer the book gave.
Where did I go wrong ?
$\frac{numerator}{denominator}$renders as $\frac{\text{numerator}}{\text{denominator}}$. It isn't that complicated. And use$\sqrt{stuff}$to get $\sqrt{\text{stuff}}$. – Xander Henderson Aug 07 '18 at 13:13$\frac{\frac{Expression_1}{Expression_2}-\frac{Expression_3}{Expression_4}}{Denominator}$yields $\frac{\frac{Expression_1}{Expression_2}-\frac{Expression_3}{Expression_4}}{Denominator}$. – Clayton Aug 07 '18 at 13:25