Finding Derivative of y = \sqrt{x + f(x^2-1)} at x = 3

[projection]

Homework Statement

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 don't understand the f(x^2-1) inside the bracket.

 

[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??

 

[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.

 

[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)

 

[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)?

 

[Mothrog]

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

 

[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}}

 

[Mothrog]

Yup. That's right.

 

[projection]

Yup.

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

 

[Mothrog]

Yes, that looks right too.

 

[projection]

Yes, that looks right too.

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?

 

[Mothrog]

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