How do you differentiate #f(x)=csc(ln(1/x)) # using the chain rule?

Question

How do you differentiate #f(x)=csc(ln(1/x))  # using the chain rule?

Charles Anctil · Accepted Answer

#d/dx(f(x))=1/xcotln(1/x)cscln(1/x)# #f(x)# composed of two functions #cscx " "and" "Ln x#. #" "# So, Differentiating #f(x)# is determined by applying chain rule. #" "# Let #u(x)=ln(1/x)" " then " "f(x)= csc(u(x))# #" "# #d/dx(f(x))=d/dx(cscx(u(x)))# #" "# #color(blue)(d/dx(f(x))=u'(x) xx csc'(u(x)))# #" "# Let us compute #color(blue)(u'(x))# #" "# #u'(x)=(color(red)((1/x)'))/(1/x)# #" "# What is the derivative of #1/x=?# #" "# The differiation of #1/x# is determined by applying the quotient rule . #" "# #color(red)((1/x)')=((1)'xxx-(x)'xx1)/x^2= (0-1)/x^2=color(red)(-1/x^2)# #" "# #color(blue)(u'(x))=(-1/x^2)/(1/x)=color(blue)(-1/x)# #" "# #" "# Knowing that the derivative of #cscx" "#is : #" "# #color(blue)((cscx)'=-cotxcscx# #" "# Let us compute #color(blue)(csc'(u(x))# #" "# #color(blue)(csc'(u(x)))=-cot(u(x))xxcsc(u(x))=color(blue)(-cotln(1/x)xxcscln(1/x)# #" "# #color(blue)(d/dx(f(x))=u'(x) xx csc'(u(x)))# #" "# #d/dx(f(x))=(-1/x)xx(-cotln(1/x)xxcscln(1/x))# #" "# Therefore, #" "# #d/dx(f(x))=1/xcotln(1/x)cscln(1/x)#

Cameron Alexander · Answer

undefined To differentiate $ f(x) = \csc(\ln(1/x)) $ using the chain rule, follow these steps:

1. Identify the outer function and the inner function. In this case, the outer function is $ \csc(x) $ and the inner function is $ \ln(1/x) $.
2. Take the derivative of the outer function with respect to the inner function, and then multiply by the derivative of the inner function with respect to $ x $.
3. The derivative of $ \csc(x) $ is $ -\csc(x) \cot(x) $.
4. The derivative of $ \ln(1/x) $ with respect to $ x $ is $ -1/x $.
5. Combine the results from steps 3 and 4 to get the final derivative.

Therefore, the derivative of $ f(x) = \csc(\ln(1/x)) $ with respect to $ x $ is:

$$ f'(x) = -\csc(\ln(1/x)) \cot(\ln(1/x)) \times \left(-\frac{1}{x}\right) $$

$$ f'(x) = \frac{\csc(\ln(1/x)) \cot(\ln(1/x))}{x} $$