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Find f'(x). Give two different solution.     F(x)=((x^2 + 1)cot(x))/(3 - cos(x)csc(x))

 Nov 1, 2014

Best Answer 

 #1
avatar+128079 
+5

Note that cosxcscx = cotx    so we have

F(x) = [(x2 + 1)*cotx] /([3 - cotx] =  [x2+1][cotx][3 - cotx]-1

And since we have three functions - call them u, v and w - we can write F(x) = u*v*w...so F '(x) = u'*v*w + u*v'*w + u*v*w'

So we have.....using the product and chain rules......

F '(x) = (2x) [x2+1][cotx][3 - cotx]-1 - [x2+1][cscx]2 [3 - cotx]-1 - [x2+1][cotx][3 - cotx]-2 [cscx]2

And we could write this in several different ways.......here's one .......

F '(x) = (2x) [x2+1][cosx][cscx][3 - cotx]-1 - [x2+1][cscx]2 [3 - cotx]-1 - [x2+1][cosx][cscx][3 - cotx]-2 [cscx]2

= (2x) [x2+1][cosx][cscx][3 - cotx]-1 - [x2+1][cscx]2 [3 - cotx]-1 + [x2+1][3 - cotx]-2[cosx][cscx]3

= [x2 + 1][cscx][3 - cotx]-2 * [(2x)][cosx][3 - cotx] - [cscx][3 - cotx] + [cosx][cscx]2]

 

 Nov 1, 2014
 #1
avatar+128079 
+5
Best Answer

Note that cosxcscx = cotx    so we have

F(x) = [(x2 + 1)*cotx] /([3 - cotx] =  [x2+1][cotx][3 - cotx]-1

And since we have three functions - call them u, v and w - we can write F(x) = u*v*w...so F '(x) = u'*v*w + u*v'*w + u*v*w'

So we have.....using the product and chain rules......

F '(x) = (2x) [x2+1][cotx][3 - cotx]-1 - [x2+1][cscx]2 [3 - cotx]-1 - [x2+1][cotx][3 - cotx]-2 [cscx]2

And we could write this in several different ways.......here's one .......

F '(x) = (2x) [x2+1][cosx][cscx][3 - cotx]-1 - [x2+1][cscx]2 [3 - cotx]-1 - [x2+1][cosx][cscx][3 - cotx]-2 [cscx]2

= (2x) [x2+1][cosx][cscx][3 - cotx]-1 - [x2+1][cscx]2 [3 - cotx]-1 + [x2+1][3 - cotx]-2[cosx][cscx]3

= [x2 + 1][cscx][3 - cotx]-2 * [(2x)][cosx][3 - cotx] - [cscx][3 - cotx] + [cosx][cscx]2]

 

CPhill Nov 1, 2014
 #3
avatar+128079 
0

LOL!!!

 

 Nov 1, 2014

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