Finding derivatives of composite functions

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crystalmath92

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Hi everybody. I'm having trouble understanding the chain rule and finding the derivatives of composite functions. If anybody can recommend any online resources to help me understand this further I would very much appreciate it. I'm trying to teach myself and my book doesn't have a very good step-by-step process in the answers section. I'll try to explain where I'm getting hung up.

The problem that I'm having trouble with is:

Find [MATH]f'(x)[/MATH] when [MATH]f(x)=x\sqrt{1-x^2}[/MATH]
1.) Using the product rule, I rewrite this as:
[MATH]f(x)=[u(x)]*[v(x)][/MATH]where,
[MATH]u(x)=x[/MATH][MATH]v(x)=\sqrt{1-x^2}[/MATH]
2.) I find the derivative of [MATH]u(x)[/MATH] and end up with:
[MATH]u'(x)=1[/MATH]
3.) I rewrite [MATH]v(x)[/MATH] into exponent notation:
[MATH]v(x)=(1-x^2)^\frac{1}{2}[/MATH]
4.) I use the power rule and attempt to find the derivative of [MATH]v(x)[/MATH]:
[MATH]v'(x)=(\frac{1}{2})(1-x^2)^\frac{-1}{2}[/MATH]
But the book is telling me that it's actually:
[MATH]v'(x)=(\frac{1}{2})(1-x^2)^\frac{-1}{2}(-2x)[/MATH]I guess I just wish that I had some more examples, or had a deeper understanding of why.
I don't know where the [MATH](-2x)[/MATH] came from.
I thought I was using the power rule correctly in this case.
ie: [MATH]n(a-x)^{n-1}[/MATH]^^Please help!^^

And then...

5.) Time to solve (using the correct derivatives) ?
[MATH]\frac{df}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}[/MATH]
[MATH]\frac{df}{dx}=x(\frac{1}{2})(1-x^2)^\frac{-1}{2}(-2x)+(1-x^2)^\frac{1}{2}(1)[/MATH]
6.) Simplify

[MATH]\frac{df}{dx}=\frac{-x^2}{\sqrt{1-x^2}}+\sqrt{1-x^2}[/MATH]
Am I going about this in the correct way?
---------------------
Thanks in advance! I really appreciate all of the help I can get.
 
crystalmath92 said:
Hi everybody. I'm having trouble understanding the chain rule and finding the derivatives of composite functions. If anybody can recommend any online resources to help me understand this further I would very much appreciate it. I'm trying to teach myself and my book doesn't have a very good step-by-step process in the answers section. I'll try to explain where I'm getting hung up.

The problem that I'm having trouble with is:

Find [MATH]f'(x)[/MATH] when [MATH]f(x)=x\sqrt{1-x^2}[/MATH]
1.) Using the product rule, I rewrite this as:
[MATH]f(x)=[u(x)]*[v(x)][/MATH]where,
[MATH]u(x)=x[/MATH][MATH]v(x)=\sqrt{1-x^2}[/MATH]
2.) I find the derivative of [MATH]u(x)[/MATH] and end up with:
[MATH]u'(x)=1[/MATH]
3.) I rewrite [MATH]v(x)[/MATH] into exponent notation:
[MATH]v(x)=(1-x^2)^\frac{1}{2}[/MATH]
4.) I use the power rule and attempt to find the derivative of [MATH]v(x)[/MATH]:
[MATH]v'(x)=(\frac{1}{2})(1-x^2)^\frac{-1}{2} * \frac{d(-2x^2)}{dx}][/MATH]......................edited​

But the book is telling me that it's actually:
[MATH]v'(x)=(\frac{1}{2})(1-x^2)^\frac{-1}{2}(-2x)[/MATH]I guess I just wish that I had some more examples, or had a deeper understanding of why.
I don't know where the [MATH](-2x)[/MATH] came from.

Your v(x) is a composite function - function of a function. It is a function of x^2 instead of x. Thus you need to apply chain rule here (in addition to the power rule) for correct work.​

I thought I was using the power rule correctly in this case.
ie: [MATH]n(a-x)^{n-1}[/MATH]^^Please help!^^

And then...

5.) Time to solve (using the correct derivatives) ?
[MATH]\frac{df}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}[/MATH]
[MATH]\frac{df}{dx}=x(\frac{1}{2})(1-x^2)^\frac{-1}{2}(-2x)+(1-x^2)^\frac{1}{2}(1)[/MATH]
6.) Simplify

[MATH]\frac{df}{dx}=\frac{-x^2}{\sqrt{1-x^2}}+\sqrt{1-x^2}[/MATH]
Am I going about this in the correct way?....................................................Yes
---------------------
Thanks in advance! I really appreciate all of the help I can get.
Click to expand...
See my comments in red above.
 
When you have to differentiate an expression like \(\displaystyle (f(x))^n\) you are using the chain rule.

Basically it boils down to doing 3 things:
1. Bring the index down the front
2. Reduce the index by 1
3. Mulitply by the derivative of the bracket

You haven't done Step 3.
 
Awesome, thanks all of you! I'm sorry, but I have another question.
The answer I came up with for the previous problem was:
[MATH]f'(x)=\frac{-x^2}{\sqrt{1-x^2}}+\sqrt{1-x^2}[/MATH]
I'm just curious, could this be simplified further?

[MATH]f'(x)=\frac{-x^2}{\sqrt{1-x^2}}+\left (\frac{\sqrt{1-x^2}}{1}\times\frac{\sqrt{1-x^2}}{\sqrt{1-x^2}}\right )[/MATH]
[MATH]f'(x)=\frac{-x^2+1-x^2}{\sqrt{1-x^2}}[/MATH]
[MATH]f'(x)=\frac{-2x^2+1}{\sqrt{1-x^2}}[/MATH]
would this be correct too? and couldn't you even rationalize the denominator after that?
 
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