How do you differentiate #y=(sin^-1x)/(1+x)#?

Question

Aurora Alvarado · Accepted Answer

#dy/dx=1/((1+x)sqrt(1-x^2)) - sin^-1x/(1+x)^2#

#y=sin^-1x/(1+x)#

Apply the quotient rule

#dy/dx= ((1+x)* d/dx sin^-1x - sin^-1x*d/dx(1+x))/(1+x)^2#

#= ((1+x)* 1/sqrt(1+x^2) - sin^-1x* 1)/(1+x)^2# (Standard differential)

#= 1/((1+x)sqrt(1-x^2)) - sin^-1x/(1+x)^2#

Scarlett Allison · Answer

undefined To differentiate the function $ y = \frac{{\sin^{-1}x}}{{1 + x}} $, you can use the quotient rule, which states that if you have a function $ y = \frac{{f(x)}}{{g(x)}} $, then its derivative is given by $ y' = \frac{{f'(x)g(x) - f(x)g'(x)}}{{[g(x)]^2}} $.

Let's denote $ f(x) = \sin^{-1}x $ and $ g(x) = 1 + x $. Then, differentiate each function individually:

1. $ f'(x) = \frac{1}{{\sqrt{1 - x^2}}} $ (derivative of $ \sin^{-1}x $ with respect to $ x $)
2. $ g'(x) = 1 $ (derivative of $ 1 + x $ with respect to $ x $)

Now, apply the quotient rule:

$$ y' = \frac{{f'(x)g(x) - f(x)g'(x)}}{{[g(x)]^2}} $$
$$ = \frac{{\frac{1}{{\sqrt{1 - x^2}}}(1 + x) - \sin^{-1}x \cdot 1}}{{(1 + x)^2}} $$
$$ = \frac{{(1 + x)}}{{(1 - x^2)^{3/2}}} - \frac{{\sin^{-1}x}}{{(1 + x)^2}} $$

So, the derivative of $ y = \frac{{\sin^{-1}x}}{{1 + x}} $ is $ y' = \frac{{(1 + x)}}{{(1 - x^2)^{3/2}}} - \frac{{\sin^{-1}x}}{{(1 + x)^2}} $.