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