Question #fb4a7

When students tell me they are stuck on problem like this I always ask, "What did you try?" -- because you can't solve challenging problems without trying something.

I think it is nice to simplify the notation by leaving the arguments out of the functions.

#r = f^4g^7/h^6#

Use the quotient and product rules along with the chain rule to find an expression for #r'#

Use the given information about the derivatives of #f#, #g# and #h# at #-9# to rewrite #r'# with no primes on the right.

Simplify.

You'll be left with a constant times #r#.

Use #r(-9) = 2# to finish.

(No, I didn't know this by looking at the question. I learned it by following the steps, because I needed to try something and each time I finished a step, I tried a next step..)

I got #72#.

This is just a very general way of applying the chain rule and quotient rule (or product rule, if that's easier here).

We know that:

• #f'(-9) = -7f(-9)#
• #g'(-9) = 10g(-9)#
• #h'(-9) = h(-9)#

So, if we have that #r(x) = ((f(x))^4(g(x))^7)/((h(x))^6)#, then by knowing that #r(-9) = 2#, we can obtain an expression in terms of #f,g,# and #h#.

#r(-9) = 2 = ((f(-9))^4(g(-9))^7)/((h(-9))^6)#,

so if we see the right-hand side, we can plug in the left-hand side. Hopefully we see this!

For now, take the derivative. Remember the chain rule after finding #[("function"(x))^n]'#.

#r'(-9) = |d/(dx)[((f(x))^4(g(x))^7)/((h(x))^6)]|_(x = 9)#

From the product rule all the way through, which may be easier:

#=> r'(x) = overbrace((f(x))^4/((h(x))^6) cdot 7(g(x))^6 cdot g'(x))^("part 1 of product rule") + underbrace((g(x))^7 cdot overbrace([(f(x))^4 cdot -6/(h(x))^(7) cdot h'(x) + 1/(h(x))^6 cdot 4(f(x))^3cdot f'(x)])^("nested product rule on" (f(x))/(h(x))))_"part 2 of product rule"#

#= (7(f(x))^4(g(x))^6)/((h(x))^6) cdot g'(x) + (g(x))^7 cdot [-(6h'(x)(f(x))^4)/(h(x))^(7) + (4(f(x))^3cdot f'(x))/(h(x))^6]#

Now, plug in the terms we know without plugging in #x = -9# yet...

#= (7r(x))/(g(x)) cdot g'(x) + (g(x))^7 cdot [-(6h'(x)(f(x))^4)/(h(x))^(7) + (4(f(x))^3cdot f'(x))/(h(x))^6]#

(note that #(r(x))/(g(x)) = ((f(x))^4(g(x))^6)/((h(x))^6)#.)

Now, at #x = -9#, we can plug in the derivative terms to remove them:

#= (7r(-9))/(g(-9)) cdot g'(-9) + (g(-9))^7 cdot [-(6h'(-9)(f(-9))^4)/(h(-9))^(7) + (4(f(-9))^3cdot f'(-9))/(h(-9))^6]#

#= (7r(-9))/(cancel(g(-9))) cdot 10cancel(g(-9)) + (g(-9))^7 cdot [-(6cancel(h(-9))(f(-9))^4)/(h(-9))^(cancel(7)^(6)) - (28(f(-9))^3cdot f(-9))/(h(-9))^6]#

Since #r(-9) = 2#, #7r(-9) cdot 10 = 7 cdot 2 cdot 10#. Distribute in #(g(-9))^7# to get:

#=> 140 + [-6stackrel(r(-9))overbrace(((g(-9))^7(f(-9))^4)/(h(-9))^6) - 28stackrel(r(-9))overbrace(((g(-9))^7(f(-9))^4)/(h(-9))^6)]#

#= 140 + [-6r(-9) - 28r(-9)]#

#= 140 + [-12 - 56]#

#= 140 - 12 - 56#

#= color(blue)(72)#