# Homework Help: Chain rule app.

1. Dec 6, 2007

### projection

1. The problem statement, all variables and given/known data

Let $$y=\sqrt{x+f(x^2-1)}$$. Find $$\frac{dy}{dx}$$ when x=3, given that $$f(8)=0$$ and $$f'(8)=3$$.

I missed the lesson for this and am lost on what to do. some guidence would be appreciated. i dont understand the $$f(x^2-1)$$ inside the bracket.

Last edited: Dec 6, 2007
2. Dec 6, 2007

### projection

i spent like an hour trying to figure this out. (need to not miss class!)

my attempt at a solution:

$$\frac{dy}{dx}=\frac{1+f'(x^2-1)}{2\sqrt{x+f(x^2-1)}}$$

$$\frac{dy}{dx}=\frac{1+f'(3^2-1))}{2\sqrt{3+f(3^2-1)}}$$

$$\frac{dy}{dx}=\frac{1+f'(3^2-1))}{2\sqrt{3+f(3^2-1)}}$$

$$\frac{dy}{dx}=\frac{1+f'(8))}{2\sqrt{3+f(8)}}$$

$$\frac{dy}{dx}=\frac{1+3}{2\sqrt{3+0}}$$

$$\frac{dy}{dx}=\frac{4}{2\sqrt{3}}$$

$$\frac{dy}{dx}=\frac{2}{\sqrt{3}}$$

this this anywhere near correct? can someone inform me where i went wrong? right of the bat? inbetween??

3. Dec 6, 2007

### Mothrog

You're very close. However, you forgot to apply the chain rule when you took the derivative of $$f(x^2 - 1)$$. That, too, is a composition of function and so you must also apply the chain rule when taking its derivative.

4. Dec 6, 2007

### projection

am not sure i understand but...would its derivative be$$\frac{1}{f(x^2-1)}$$? or could t be $$f'(x^2-1)(2x)$$

Last edited: Dec 6, 2007
5. Dec 6, 2007

### Mothrog

Not quite. Remember that the chain rule says

$$\frac{d}{dx}f(g(x))=f'(g(x))g'(x)$$​

When you have

$$f(x^2 - 1)$$​

They're saying that, rather than plugging in the usual x in your variable, you plug in $$x^2 - 1$$. For example, let $$f(t) = t^2$$. Then, $$f(x^2) = (x^2)^2 = t^4$$. So, you have a composition of functions, where, using the notation of the chain rule as stated above,

$$f(x) = f(x)$$

$$g(x) = x^2 - 1$$​

So, thinking about the chain rule, what would the derivative with respect to x be of $$f(x^2 - 1)$$?

Last edited: Dec 6, 2007
6. Dec 6, 2007

### Mothrog

Yes, the correct answer is $$f'(x^2-1)(2x)$$.

7. Dec 6, 2007

### projection

so new solution is:

$$\frac{dy}{dx}=\frac{1+f'(x^2-1)(2x)}{2\sqrt{x+f(x^2-1)}}$$

$$\frac{dy}{dx}=\frac{1+f'(8)(2(3))}{2\sqrt{3+f(8)}}$$

$$\frac{dy}{dx}=\frac{1+18}{2\sqrt{3}}$$

$$\frac{dy}{dx}=\frac{19}{2\sqrt{3}}$$

8. Dec 6, 2007

### Mothrog

Yup. That's right.

9. Dec 6, 2007

### projection

great. thanks a lot man. really starting to understand.

another problem. i have soloved it, but need to see if i have done it correctly.

Q-skydiver jumps from plane at 3000m. distance fallen in meters after t seconds is:

$$s=5t^2$$

During fall, experiences air pressure $$p$$ that will cause his ears to pop if the rate of change of pressure $$\frac{dp}{dt}$$ exceeds 2 pressure units/s. suppose that the rate of change of pressure with respect to distance fallen in metres is 0.075 pressure units/m. what time will the sky divers ears pop? At what height will this occur?

Solution attempt:
A)time ears pop

$$\frac{dp}{dt}=\frac{dp}{ds}*\frac{ds}{dt}$$

$$\frac{dp}{dt}=2$$

$$\frac{dp}{ds}=0.075$$

$$\frac{ds}{dt}= 10t$$

$$2=0.075*10t$$

$$t=2.7$$

b)height of poping

$$s=5(2.7)^2$$
$$s=35.6$$
$$3000-35.6=2964.4$$

10. Dec 6, 2007

### Mothrog

Yes, that looks right too.

11. Dec 6, 2007

### projection

cool. that's 2 out of 4 on the worksheet. onto number three.

suppose that f and g are functions such that:

$$f(1)=-\frac{1}{2}$$ , $$f'(1)=-\frac{2}{3}$$ , $$g(2)=1$$ and $$g'(2)=3$$

find $$h'(2)$$ where $$h$$ is the composite fuction $$h(x)=f(g(x))$$.

so in this to get $$h'(2)$$ i need to just take the derivative $$f'(1)$$ and multiply it by the derivative $$g'(2)$$ ?

$$-\frac{2}{3} * 3$$.

so $$h'(2)$$ is equal to -2?

it seems wrong to me because i have done it too simply. did i completely miss the mark? or partialy?

12. Dec 6, 2007

### Mothrog

I think you understand the chain rule, so you should have a good idea whether that is correct or not.