Calculus: Composite and Quotient rules. HELP

[Jonathanjc]

Homework Statement

Remember to show your working explicitly throughout your answer to this question.

(a) (i) Use the Composite Rule to differentiate the function f(x) = (x^2− 6x + 23)^(3/2)

(ii) Use the Quotient Rule and your answer to part (a)(i) to show that the function:
g(x) = (x − 2)/(x^2 − 6x + 23)^(3/2)

has derivative

g'(x) = (5 + 9x − 2x^2)/(x2 − 6x + 23)^(5/2)

Homework Equations

(a)(i) Composite Rule (Leibniz form) (Chain Rule)
If y = g(u), where u = f(x), then
dy/dx=(dy/du)(du/dx)

(ii) Quotient Rule (Leibniz form)
If y = u/v, where u = f(x) and v = g(x), then

dy/dx = 1/v^2(v(du/dx)-u(dv/dx)

The Attempt at a Solution

Using the leibniz form of the composite rule, otherwise known as the chain rule:

We have: d/dx(f(x)) = d/dx((x^2-6x+23)^(3/2))

- The derivative of f(x) is f'(x)
=f'(x) =d/dx((x^2-6x+23)^(3/2))

-Using the chain rule:
d/dx((x^2-6x+23)^(3/2))=(du^(3/2)/du)(du/dx)

Where, u = x^2-6x+23, and

(du^(3/2)/du)(3(sqrt)u/2)

=f'(x) = 3/2(sqrtx^2-6x+23)(d/dx(x^2-6x+23))

= f'(x) = 3/2(sqrtx^2-6x+23)(2x-6)

My problem now is that for part (ii) the f'(x) is not in the form for the quotient rule and I am having trouble re-arranging it to get the appropriate form.

Can you please enlighten me as I have been staring at it for hrs with no concept in how to approach it.

 

[Deadstar]

g(x) = (x − 2)/(x^2 − 6x + 23)^(3/2)

u = (x-2)

v = (x^2 − 6x + 23)^(3/2)

I can't really understand your problem here... What do mean when you say "the f'(x) is not in the form for the quotient rule" ??

EDIT: Hold on.. Are you getting confused since they used f(x) and g(x) for a few things.

 

[Jonathanjc]

I hope not, what I mean is that for part (i) I need to differentiate the function, f(x) by using the chain rule, which is f'(x) and is my solution:

f'(x) = 3/2(sqrtx^2-6x+23)(2x-6)

So, as far as I can see part (ii) says that my solution above has to be used with the quotient rule to find the new function of g(x) and then g'(x) but the solution I have found isn't in the form of the normal quotient rule:

k(x) = f(x)/g(x)

I hope that makes it clearer.

J

 

[Deadstar]

Not really...

You have u=(x-2) and v=(x^2 − 6x + 23)^(3/2)

y = g(x) = (x − 2)/(x^2 − 6x + 23)^(3/2)

dy/dx = (1/v^2)(v(du/dx)-u(dv/dx))

But from part 1) you know what dv/dx is (3/2(sqrtx^2-6x+23)(2x-6))

you know what v and u are, just find du/dx then plug everything into the equation (1/v^2)(v(du/dx)-u(dv/dx)).

I think you're thinking you need to use the quotient rule on (x^2 − 6x + 23)^(3/2) but why would you need to do that? Read the question again.

Use the Quotient Rule AND your answer to part (a)(i) to show that the function:
g(x) = (x − 2)/(x^2 − 6x + 23)^(3/2)

has derivative...


I.E. Use the quotient rule on the above equation.

 

[Jonathanjc]

Oh, so not turn the solution of part ai into part aii.

Ok, I think I got it.

Thanks

 

[HallsofIvy]

Use the result of part ai in part aii :
the derivative of g(x)= (x − 2)/(x^2 − 6x + 23)^(3/2)= (x- 2)/f(x)
g'= [(x- 2)'f(x)- (x- 2)f'(x)]/f^2(x)= [f(x)- (x-2)f'(x)]/f^2(x)

ai is f'(x).