Chain Rule Problem

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March 20th 2010, 10:53 AM
A Beautiful Mind

Chain Rule Problem

$sin^2(x^2) = sin(x^2)cos(x^2)$

I need to know how to get this part.

Here's the problem:

$f(x) = sin^2(x)sin(x^2)sin^2(x^2)$

Now it's obvious that you can use the product rule to differentiate the three of these, which I've done.

It'll look something like this:

$f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x)$

However, on the last part, I'm only able to get 2 parts of the answer. The $f(x)$ and $g(x)$ are kept constant while I'm having trouble differentiating the $h'(x)$ :

$sin^2(x^2)$

When seemed obvious to me was that you'd bring the power of 2 to the outside and then differentiate, which would lead to $2sin(x^2)cos(x^2)2x,$ but this is not the answer in the back of the book.

$f(x)g(x)h'(x)$ is supposed to be = $sin(x^2)cos(x^2)sin^2xsinx^2$

I had four tutors working on this problem who gave up on it at my college, which I thought was either sad or Michael Spivak (author of this legendary Calculus text) made an error.
March 20th 2010, 11:08 AM
harish21

Quote:

Originally Posted by A Beautiful Mind

$sin^2(x^2) = sin(x^2)cos(x^2)$

I need to know how to get this part.

Here's the problem:

$f(x) = sin^2(x)sin(x^2)sin^2(x^2)$

Now it's obvious that you can use the product rule to differentiate the three of these, which I've done.

It'll look something like this:

$f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x)$

However, on the last part, I'm only able to get 2 parts of the answer. The $f(x)$ and $g(x)$ are kept constant while I'm having trouble differentiating the $h'(x)$ :

$sin^2(x^2)$

When seemed obvious to me was that you'd bring the power of 2 to the outside and then differentiate, which would lead to $2sin(x^2)cos(x^2)2x,$ but this is not the answer in the back of the book.

$f(x)g(x)h'(x)$ is supposed to be = $sin(x^2)cos(x^2)sin^2xsinx^2$

I had four tutors working on this problem who gave up on it at my college, which I thought was either sad or Michael Spivak (author of this legendary Calculus text) made an error.

$f(x) = sin^2(x)sin(x^2)sin^2(x^2)$

so, $f(x) = sin^2(x) \times sin^3(x^2)$

Wont it be easier now to differentiate $f(x)$ to get:

$f'(x)= 2 sin(x) sin^3(x^2) cos(x)+6 x sin^2(x) sin^2(x^2) cos(x^2)$
March 20th 2010, 11:09 AM
coscos

I believe it will be easier if you start by writing it as sin^2(x)*sin^4(x^2). You know that the derivative of f(x)g(x) = f'(x)g(x)+f(x)g'(x). f(x)=sin^2(x) f'(x)=2sin(x)cos(x).
For sin^4(x^2) is easier if you for instance set u=sin(x^2). g(u)=u^4 g'(u)=4*u^3*u'. now determinate u'. u=sin(x^2), u'=cos(x^2)*2x. you can now determinate that g'(x)=4*u^3*u'= 4*(sin(x^2))^3*2xcos(x^2).

I hope that this notation-thing will help you as it has helped me, it's much easier to have control when facing problems involvning derivatives of functions with inner derivatives, who might themselves have inner derivatives etc etc!
March 20th 2010, 11:13 AM
A Beautiful Mind

I'll show you what the answer is supposed to look like. There wasn't any condensing.

$(2sinxcosxsinx^2sin^2x^2) + (2xcosx^2sin^2xsin^2x^2)+(sinx^2cosx^2sin^2xsinx^2 )$
March 20th 2010, 11:27 AM
harish21

Quote:

Originally Posted by A Beautiful Mind

I'll show you what the answer is supposed to look like. There wasn't any condensing.

$(2sinxcosxsinx^2sin^2x^2) + (2xcosx^2sin^2xsin^2x^2)+(sinx^2cosx^2sin^2xsinx^2 )$

The first two terms I got are the same as yours. However the third term of my answer looks like this:

$4x \times sin(x^2) \times cos(x^2) \times sin^2(x) \times sin(x^2)$