Differentiation using the Composite Rule

**honeybumble** · September 15th 2008, 11:21 AM

Hi, I'm new to this forum and am abit stuck.

I need to differentiate the function f(x)=10+x(squared) - this function is also a square root but I dont know how to input that here.

What I have so far is y=sq.root u, where u = 10+x(squared)
then dy/du = 1/2 10/sq.rootu and du/dx=2x
I think the composite rule says dy/dx = dy/du du/dx = 1/2 1/sq.root u. 2x = x/sq.root 10+x(squared). Hopefully this makes sense!?

I then have to use the quotient rule together with my answer above to show that the function g(x)= (e to the power of x/7)/sq.root 10 + x(squared) has the derivative

g'(x) = e to the power of x/7(xsquard - 7x + 10) divided by 7(10 + x squared) to the power of 3/2
I will be extremely grateful for any help.
Thanks

**icemanfan** · September 15th 2008, 11:33 AM

Your first differentiation is correct. The derivative of $\sqrt{10 + x^2}$ is indeed $\frac{x}{\sqrt{10 + x^2}}$ .

For the second problem, the derivative of $\frac{f(x)}{g(x)}$ is $\frac{f(x)g'(x) - g(x)f'(x)}{(g(x))^2}$ .

**Moo** · September 15th 2008, 11:37 AM

Originally Posted by honeybumble

Hi, I'm new to this forum and am abit stuck.

I need to differentiate the function f(x)=10+x(squared) - this function is also a square root but I dont know how to input that here.

$f^2(x)=10+x^2$
If you don't want to use latex, use ^ for exponents, and parentheses please

What I have so far is $y=\sqrt{u}$ , where $u=10+x^2$
then $\frac{dy}{du} =\frac 12 \times \frac{1}{\sqrt{u}}$ and $\frac{du}{dx}=2x$

Correct so far

I think the composite rule says $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$

Yes it does !

$\frac 12 \times \frac{1}{\sqrt{u}} \times 2x=\frac{x}{\sqrt{10+x^2}}$ . Hopefully this makes sense!?

Yes.

Now, a problem with that. If $f^2(x)=10+x^2$ then $f(x)=\sqrt{10+x^2}$ OR $f(x)=-\sqrt{10+x^2}$ !!!

**honeybumble** · September 15th 2008, 01:05 PM

Thanks for your help! It is much appreciated.

I have further problems I would be grateful of further clarification.

Firstly, how do I find any stationary points of the function g(x) as defined by the quotient rule component of my previous post. I also need to use the First Derivative Test to classify each stationary point as a local maximum or local minimum of g(x).

Also what are the rules to find the general solution of the differential equation (using the answer to the f(x) function in my first post) of the differential equation

dy/du = 10+y(squared) / y (x>0, y>0) - sorry I cant work out where to find the "squared" and "square root" symbols on this page....

giving the answer in implicit form

and then find the solution of the differential equation for which y + sq.root6 when x=0 giving this particular solution in explicit form.

Many thanks

**11rdc11** · September 15th 2008, 01:23 PM

To find your stationary points equal your derivative to 0 and see what your x values are. To use the first derivative test you then have to see in what interval the sign changes from positive to negative or vice versa. Here is an example, let me know if you need me to explain some more.

http://www.mathhelpforum.com/math-he...tive-test.html

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