How do you differentiate #f(x) = (1-sqrt(3x-1))^2 # using the chain rule?

Question

How do you differentiate #f(x) = (1-sqrt(3x-1))^2     # using the chain rule?

Isabella Ackerman · Accepted Answer

#y = u^2#

#u = (1 - sqrt(3x - 1))#

The chain rule states #dy/dx = dy/(du) xx (du)/dx#.

We must therefore differentiate both functions.

#y' = 2u#

#u# is a little more complicated: we will have to apply the chain rule to the radical.

The derivative of any constant #c# is always #0#, so we don't have to worry about the #1#, thankfully.

#u = -v^(1/2)#

#v = 3x - 1#

#u' = -1/(2v^(1/2))#

#v' = 3#

#dy/dx = dy/(du) xx (du)/dx#

#dy/dx = 3 xx -1/(2v^(1/2))#

#dy/dx = -3/(2v^(1/2))#

Now that we know the derivative of #u#, we can use the chain rule again to determine the derivative of #f(x)#.

#f'(x) = -3/(2v^(1/2)) xx 2u#

#f'(x) = -3/(2(3x - 1)^(1/2)) xx 2(1 - sqrt(3x - 1))#

#f'(x) = -3/(2(3x - 1)^(1/2)) xx 2 - 2sqrt(3x - 1)#

#f'(x) = (-3(2 - 2sqrt(3x - 1)))/(2(3x - 1)^(1/2)#

#f'(x) = (-6 + 6sqrt(3x- 1))/(2sqrt(3x - 1))#

#f'(x) = (-6(1 - 1sqrt(3x - 1)))/(2sqrt(3x - 1)#

#f'(x) = (-3(1 - 1sqrt(3x - 1)))/(sqrt(3x - 1)#

#f'(x) = (-3 + 3sqrt(3x- 1))/(sqrt(3x - 1)#

#f'(x) = (3sqrt(3x - 1) - 3)/(sqrt(3x - 1)#

It's long, but it works!

Hopefully you understand this complicated problem better now!

Charles Arocha · Answer

undefined To differentiate $ f(x) = (1 - \sqrt{3x - 1})^2 $ using the chain rule, follow these steps:

1. Identify the outer function: $ u = (1 - v)^2 $, where $ v = \sqrt{3x - 1} $.
2. Differentiate the outer function with respect to its inner variable: $ \frac{du}{dv} = 2(1 - v) $.
3. Identify the inner function: $ v = \sqrt{3x - 1} $.
4. Differentiate the inner function with respect to $ x $: $ \frac{dv}{dx} = \frac{1}{2\sqrt{3x - 1}} $.
5. Apply the chain rule: $ \frac{df}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} $.
6. Substitute the expressions for $ \frac{du}{dv} $ and $ \frac{dv}{dx} $ into the chain rule formula.
7. Simplify the expression.

$ \frac{df}{dx} = 2(1 - \sqrt{3x - 1}) \cdot \frac{1}{2\sqrt{3x - 1}} $

$ \frac{df}{dx} = \frac{1 - \sqrt{3x - 1}}{\sqrt{3x - 1}} $