How do you find the derivative of #f(x)=csc(3x-1)#?

Question

Scarlett Allison · Accepted Answer

#f'(x)=3cos(3x-1)csc^2(3x-1)# #f(x)# is a xomposite of two functions Let : #color(blue)(g(x)=3x-1) and color(brown)(h(x)=cscx # So, #f(x)=color(brown)hcolor(blue)((g(x))# so thederivative of this function is by applying chain rule: #color(red)(f'(x))=color(red)(h'(g(x)*(g'(x))# #color(red)h'(g(x))=????# #color(brown)(h(x)=cscx # #h'(x)=cosx*csc^2x# #color(red)h'(g(x))=cos(g(x))*csc^2(g(x))=cos(3x-1)*csc^2(3x-1)# #color(red)(h'(g(x))) = cos(3x-1)*csc^2(3x-1)# #color(red)(g'(x))=????# #color(blue)(g(x)=3x-1)# #color(red)(g'(x))=3# #color(red)(f'(x))=color(red)(h'(g(x)*(g'(x))# #color(red)(f'(x))=cos(3x-1)*csc^2(3x-1)*3# #color(red)(f'(x)=3cos(3x-1)csc^2(3x-1))#

Aurora Alvarado · Answer

undefined To find the derivative of $ f(x) = \csc(3x - 1) $, you can use the chain rule. The derivative of $ \csc(u) $ with respect to $ u $ is $ -\csc(u) \cot(u) $. So, using the chain rule, the derivative of $ f(x) $ with respect to $ x $ is:

$$ f'(x) = -\csc(3x - 1) \cot(3x - 1) \cdot \frac{d}{dx}(3x - 1) $$

$$ = -\csc(3x - 1) \cot(3x - 1) \cdot 3 $$

$$ = -3\csc(3x - 1) \cot(3x - 1) $$

So, the derivative of $ f(x) $ is $ -3\csc(3x - 1) \cot(3x - 1) $.