# Thread: Plz help ASAP chain rule and rates

1. ## Plz help ASAP chain rule and rates

Find h' in terms of f' and g' if h(x) = f(g(cos(4x5+lnx)))

I know h'(x) = f'g(x) . g'(x)

and I have g'(x) = -(20x4 + 1/x)sin(4x5+lnx) but does the f'g(x) mean that I must double differentiate or am i just interpreting it wrong and that's the answer?

y = ln (x.sqrt(x3+2)/tanx) find dy/dx if possible. I would like to know if it exists and if so what conditions there are. I know for

y = ln(9-x2) - ln(x-4) no differential exists due to the conditions of -3<x<3 and x>4

I also have no idea how to answer the following either:

Two cars, A and B, start moving from the same point P. Car A travels south at 60 km/h, while
Car B (the very slow driver!) travels west at 25 km/h. At what rate is the distance between the
cars increasing two hours later?

About how accurately should we measure the radius r of a sphere to calculate the surface area
S = 4 pi r2 to within 1% of its true value?

2. ## Re: Plz help ASAP chain rule and rates

It's $h'(x) = f'(g(x)) \cdot g'(x)$. Essentially, you treat g(x) as if it were a variable for the first part, then multiply by the derivative of g. For your problem, you have to do the chain rule twice, since you have f(g(a third function)).

For the second one, I think you just calculate the derivative and look at the formula to see where it's undefined (no 0 in denominator, no square root of negative, no ln of negative or zero).

For the third, I think your analysis is correct. You can combine the logarithms to get it to be defined, but as it's written the function is undefined everywhere.

For the fourth, try calculating the distance after 1 hour.

And for the last, you can find dS/dr. Use this as an estimate of (error in S)/(error in r). But the % error in S means (error in S)/S. So

$\text{percent error in S} = \frac {\text{error in S}}{S} = \frac{1}{S} \frac{dS}{dr} \text{(error in r)} = \frac{1}{S} \frac{dS}{dr} r \text{(percent error in r)}$.

Hollywood

3. ## Re: Plz help ASAP chain rule and rates

For problem 4:
You can use related rates.
Let A= Distance covered by car A
B = Distance covered by car B
X = Distance between A and B

A, B and X form a right angled triangle where:

X² = A² + B²

Use distance = speed * time
to find A = 25*2=50
B=60*2=120

X² = 50² + 120²
X=130

distance/time = speed
130/2 = 65km/h