How do you find the derivative of #arctan [(1-x)/(1+x)]^(1/2)#?

Question

Aurora Alvarado · Accepted Answer

#d/dx(tan^-1 sqrt((1-x)/(1+x)))=-1/2*(sqrt(1-x^2))/((1-x^2))#

Formula for finding derivative of arctangent

#d/dx(tan^-1 u)=(1/(1+u^2))d/dx(u)#

#d/dx(tan^-1 sqrt((1-x)/(1+x)))=(1/(1+(sqrt((1-x)/(1+x)))^2))d/dx(sqrt((1-x)/(1+x)))#

#d/dx(tan^-1 sqrt((1-x)/(1+x)))=# #(1/(1+(1-x)/(1+x)))(1/(2(sqrt((1-x)/(1+x)))))*d/dx((1-x)/(1+x))#

#d/dx(tan^-1 sqrt((1-x)/(1+x)))=# #(1/(1+(1-x)/(1+x)))(1/(2(sqrt((1-x)/(1+x)))))*(((1+x)(-1)-(1-x)(1))/(1+x)^2)#

Simplify

#d/dx(tan^-1 sqrt((1-x)/(1+x)))=# #((1+x)/(1+x+1-x))(1/2*sqrt((1+x)/(1-x)))*((-1-x-1+x)/(1+x)^2)#

#d/dx(tan^-1 sqrt((1-x)/(1+x)))=# #((1+x)/(2))(1/2*sqrt((1+x)/(1-x)))*((-2)/(1+x)^2)#

#d/dx(tan^-1 sqrt((1-x)/(1+x)))=# #(1/2*sqrt((1+x)/(1-x)))*((-1)/(1+x))#

#d/dx(tan^-1 sqrt((1-x)/(1+x)))=# #(1/2*sqrt(((1+x)/(1-x))((1-x)/(1-x))))*((-1)/(1+x))#

#d/dx(tan^-1 sqrt((1-x)/(1+x)))=-1/2*(sqrt(1-x^2))/((1-x^2))#

God bless....I hope the explanation is useful.

Charles Anctil · Answer

undefined To find the derivative of $ \arctan\left[\left(\frac{1-x}{1+x}\right)^{\frac{1}{2}}\right] $, you can use the chain rule.

Let $ u = \left(\frac{1-x}{1+x}\right)^{\frac{1}{2}} $. Then differentiate $ u $ with respect to $ x $ using the chain rule, and you'll get:

$$ \frac{du}{dx} = \frac{1}{2}\left(\frac{1}{1+x}\right)\left(\frac{-1-1}{(1+x)^2}\right)\left(\frac{1-x}{1+x}\right)^{-\frac{1}{2}} $$

Now, differentiate $ \arctan(u) $ with respect to $ u $, and then multiply by $ \frac{du}{dx} $:

$$ \frac{d}{dx} \left[\arctan\left(\left(\frac{1-x}{1+x}\right)^{\frac{1}{2}}\right)\right] = \frac{1}{1+u^2} \cdot \frac{du}{dx} $$

Substitute $ u $ and $ \frac{du}{dx} $ back into the equation to get the derivative:

$$ \frac{d}{dx} \left[\arctan\left(\left(\frac{1-x}{1+x}\right)^{\frac{1}{2}}\right)\right] = \frac{1}{1+\left(\left(\frac{1-x}{1+x}\right)^{\frac{1}{2}}\right)^2} \cdot \frac{1}{2}\left(\frac{1}{1+x}\right)\left(\frac{-1-1}{(1+x)^2}\right)\left(\frac{1-x}{1+x}\right)^{-\frac{1}{2}} $$

Simplify this expression to get the final derivative.

Emilia Aguilar · Answer

undefined To find the derivative of $ \arctan \left[ \left( \frac{1-x}{1+x} \right)^{\frac{1}{2}} \right] $, you can use the chain rule. Let $ u = \left( \frac{1-x}{1+x} \right)^{\frac{1}{2}} $. Then the function becomes $ \arctan(u) $.

Now, apply the chain rule, which states that if $ y = f(g(x)) $, then $ y' = f'(g(x)) \cdot g'(x) $.

$$ \frac{dy}{dx} = \frac{d}{du} \left( \arctan(u) \right) \cdot \frac{du}{dx} $$

$$ = \frac{1}{1 + u^2} \cdot \frac{d}{dx} \left( \left( \frac{1-x}{1+x} \right)^{\frac{1}{2}} \right) $$

$$ = \frac{1}{1 + \left( \frac{1-x}{1+x} \right)} \cdot \frac{d}{dx} \left( \left( \frac{1-x}{1+x} \right)^{\frac{1}{2}} \right) $$

$$ = \frac{1}{1 + \left( \frac{1-x}{1+x} \right)^2} \cdot \frac{d}{dx} \left( \left( \frac{1-x}{1+x} \right)^{\frac{1}{2}} \right) $$

Now, you can use the chain rule again to find $ \frac{d}{dx} \left( \left( \frac{1-x}{1+x} \right)^{\frac{1}{2}} \right) $. Let $ v = \frac{1-x}{1+x} $, then the function becomes $ v^{\frac{1}{2}} $.

$$ \frac{d}{dx} \left( \left( \frac{1-x}{1+x} \right)^{\frac{1}{2}} \right) = \frac{d}{dv} (v^{\frac{1}{2}}) \cdot \frac{d}{dx} \left( \frac{1-x}{1+x} \right) $$

$$ = \frac{1}{2} v^{-\frac{1}{2}} \cdot \frac{d}{dx} \left( \frac{1-x}{1+x} \right) $$

$$ = \frac{1}{2} \left( \frac{1-x}{1+x} \right)^{-\frac{1}{2}} \cdot \frac{d}{dx} \left( \frac{1-x}{1+x} \right) $$

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

$$ = \frac{1}{2} \left( \frac{2x}{(1+x)^2} \right) $$

Putting it all together:

$$ \frac{dy}{dx} = \frac{1}{1 + \left( \frac{1-x}{1+x} \right)^2} \cdot \frac{1}{2} \left( \frac{2x}{(1+x)^2} \right) $$

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