How do you use the chain rule to differentiate #y=(x+1)^6/(3x-2)^5#?

Question

Paisley Johnson · Accepted Answer

You avoid the quotient rule, and take the derivative using the product rule. The chain rule states:

#(df)/(dx) = (df)/(dy)(dy)/(dx)#

So, you incorporate the nested function, and take the derivative of that function as well.

#color(blue)((dy)/(dx)) = (x+1)^6 * d/(dx)[1/(3x-2)^5] + 1/(3x-2)^5*d/(dx)[(x+1)^6]#

#= (x+1)^6 * -(5)/(3x-2)^6*d/(dx)[3x-2] + 1/(3x-2)^5 * 6(x+1)^5 * d/(dx)[x+1]#

#= color(blue)(-(15(x+1)^6)/(3x-2)^6 + (6(x+1)^5)/(3x-2)^5)#

Charles Anctil · Answer

undefined To differentiate $ y = \frac{(x+1)^6}{(3x-2)^5} $ using the chain rule, follow these steps:

1. Identify the inner and outer functions.
   - Inner function: $ u = x + 1 $
   - Outer function: $ y = u^6 $

2. Compute the derivatives of the inner and outer functions.
   - $ \frac{du}{dx} = 1 $
   - $ \frac{dy}{du} = 6u^5 $

3. Apply the chain rule:
   - $ \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} $

4. Substitute the derivatives and the inner function into the chain rule formula:
   - $ \frac{dy}{dx} = 6u^5 \times 1 $

5. Replace $ u $ with the inner function $ x + 1 $:
   - $ \frac{dy}{dx} = 6(x + 1)^5 $

6. Repeat the process for the denominator.
   - Inner function: $ v = 3x - 2 $
   - Outer function: $ y = v^{-5} $

7. Compute the derivatives of the inner and outer functions.
   - $ \frac{dv}{dx} = 3 $
   - $ \frac{dy}{dv} = -5v^{-6} $

8. Apply the chain rule:
   - $ \frac{dy}{dx} = \frac{dy}{dv} \times \frac{dv}{dx} $

9. Substitute the derivatives and the inner function into the chain rule formula:
   - $ \frac{dy}{dx} = -5v^{-6} \times 3 $

10. Replace $ v $ with the inner function $ 3x - 2 $:
    - $ \frac{dy}{dx} = -15(3x - 2)^{-6} $

Therefore, the derivative of $ y = \frac{(x+1)^6}{(3x-2)^5} $ with respect to $ x $ is:
$$ \frac{dy}{dx} = 6(x + 1)^5 \cdot -15(3x - 2)^{-6} $$