Chain Rule Problem

[ChubbyBoi]

What am I doing wrong here? I have applied the chain rule, but the answer is actually 92/9.

 

[DrPhil]

What am I doing wrong here? I have applied the chain rule, but the answer is actually 92/9.

[I always use the "prime" notation to indicate the derivative with respect to the independent variable of the function .. that is, y' indicates dy/du, and u'=du/dx. To avoid confusion, lets bypass the primes.]

The error I see is that the "3" from dy/du must also be multiplied by du/dx:

\(\displaystyle \displaystyle \dfrac{dy}{dx} = \dfrac{dy}{du}\times \dfrac{du}{dx} = (10u + 3)\times \dfrac{du}{dx}\)

.....\(\displaystyle \displaystyle = \left(10\dfrac{18}{x^2 + 5} + 3\right)\times \left(\dfrac{-36x}{(x^2 + 5)^2} \right) \)

When you evaluate at x=2, will that give you the right answer?

 

[ChubbyBoi]

[I always use the "prime" notation to indicate the derivative with respect to the independent variable of the function .. that is, y' indicates dy/du, and u'=du/dx. To avoid confusion, lets bypass the primes.]

The error I see is that the "3" from dy/du must also be multiplied by du/dx:

\(\displaystyle \displaystyle \dfrac{dy}{dx} = \dfrac{dy}{du}\times \dfrac{du}{dx} = (10u + 3)\times \dfrac{du}{dx}\)

.....\(\displaystyle \displaystyle = \left(10\dfrac{18}{x^2 + 5} + 3\right)\times \left(\dfrac{-36x}{(x^2 + 5)^2} \right) \)

When you evaluate at x=2, will that give you the right answer?

Thank you for spotting that stupid mistake, I thought that when the derivative of a composite function equates to 1 meaning the power of u goes to 0, than we do not have to multiply by the derivative of u. That was the mistake, however I still do not get the answer of 92/9, instead I get -184/9 as most of my friends are getting.

 

[JeffM]

What am I doing wrong here? I have applied the chain rule, but the answer is actually 92/9.

Let's try using substitutions all the way through

\(\displaystyle w = x^2 + 5 \implies \dfrac{dw}{dx} = 2x.\)

\(\displaystyle u = \dfrac{18}{x^2 + 5} = \dfrac{18}{w} = 18w^{-1} \implies \dfrac{du}{dw} = -18w^{-2} = \dfrac{-18}{w^2} = \dfrac{-18}{(x^2 + 5)^2} \implies\)

\(\displaystyle \dfrac{du}{dx} = \dfrac{du}{dw} * \dfrac{dw}{dx} = \dfrac{-18}{(x^2 + 5)^2} * 2x = \dfrac{-36x}{(x^2 + 5)^2}.\)

\(\displaystyle y = 5u^2 + 3u - 1 \implies \dfrac{dy}{du} = 10u + 3 = \dfrac{180}{x^2 + 5} + 3 = \dfrac{180 + 3x^2 + 15}{x^2 + 5} = \dfrac{3x^2 + 195}{x^2 + 5} =\dfrac{3(x^2 + 65)}{x^2 + 5} \implies\)

\(\displaystyle \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dx} = \dfrac{3(x^2 + 65)}{x^2 + 5} * \dfrac{-36x}{(x^2 + 5)^2} = \dfrac{-108x(x^2 + 65)}{(x^2 + 5)^3}.\)

I think the answer in the answer key is incorrect.

\(\displaystyle \dfrac{-108 * 2(2^2 + 65)}{(2^2 + 5)^3} = \dfrac{-216(4 + 65)}{(4 + 5)^3} = \dfrac{-9 * 24 * 69}{9^3} = \dfrac{- 3 * 8 * 3 * 23}{9^2} = \dfrac{- 8 * 23}{9} = \dfrac{- 184}{9}.\)