# Thread: Cannot work out application of chain rule

1. ## Plz help ASAP cannot work out application of chain rule and few others

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: Cannot work out application of chain rule

Hey yob007.

The f'(g(x)) just means f'(g(cos(4x5+lnx))).

For the other two, just differentiate the expressions and show when the derivative doesn't exist (either from the domain or when the derivative can't be computed).

For the last set of questions, show us what you have tried and what you have thought about: doesn't matter if its a partial answer or an incomplete brainstorm.

3. ## Re: Cannot work out application of chain rule

Does that mean I differentiate -(20x4 + 1/x)sin(4x5+lnx) and multiply it by g'(x)?

4. ## Re: Cannot work out application of chain rule

You are actually going to have a do the chain rule twice on this one (sorry I wasn't on the ball).

Your function looks like h(x) = f(g(b(x))) where b(x) = cos(4x5+lnx). So differentiating this gives:

h'(x)
= f'(x)
= d/dx(g(b(x))) * f'(g(b(x)))
= b'(x) * g'(b(x)) * f'(g(b(x))

So you'll get three terms for this one.