How do you use the chain rule to differentiate #y=(x+1)^(-1/2)#?

Question

Isabella Ackerman · Accepted Answer

The answer is #=-1/2(x+1)^(-3/2)# The function is #=u^(-1/2)# We use #(x^n)'=nx^(n-1)# The derivative is #=(u^(-1/2))'=-1/2u^(-3/2)*u'# Here, #y=(x+1)^(-1/2)# #dy/dx=-1/2(x+1)^(-3/2)*1#

Emilia Aguilar · Answer

undefined To differentiate $ y = (x + 1)^{-1/2} $ using the chain rule:

1. Identify the outer function, which is $ f(u) = u^{-1/2} $, and the inner function, which is $ g(x) = x + 1 $.
2. Find the derivative of the outer function with respect to its variable. The derivative of $ f(u) $ with respect to $ u $ is $ f'(u) = -\frac{1}{2}u^{-3/2} $.
3. Find the derivative of the inner function with respect to its variable. The derivative of $ g(x) $ with respect to $ x $ is $ g'(x) = 1 $.
4. Apply the chain rule, which states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.
5. Substitute the inner function $ u = g(x) = x + 1 $ into $ f'(u) $.
6. Multiply $ f'(u) $ by $ g'(x) $ to obtain the derivative of the composite function $ y $ with respect to $ x $.

The derivative of $ y $ with respect to $ x $ is:

$$ \frac{dy}{dx} = f'(u) \cdot g'(x) = -\frac{1}{2}(x + 1)^{-3/2} \cdot 1 = -\frac{1}{2(x + 1)^{3/2}} $$

So, $ \frac{dy}{dx} = -\frac{1}{2(x + 1)^{3/2}} $.