How do you differentiate #f(x)=csc(sqrt(e^x)) # using the chain rule?

Question

How do you differentiate #f(x)=csc(sqrt(e^x))  # using the chain rule?

Charles Anctil · Accepted Answer

#f'(x)=(-sqrt(e^x)csc(sqrt(e^x))cot(sqrt(e^x)))/2#

Use the chain rule (a lot of times!):

First issue: the cosecant function: #d/dx(csc(u))=-csc(u)cot(u)*u'#

#f'(x)=-csc(sqrt(e^x))cot(sqrt(e^x))*d/dx(sqrt(e^x))#

Second issue: the square root: #d/dx(sqrtu)=d/dx(u^(1/2))=1/2u^(-1/2)*u'#

#f'(x)=-csc(sqrt(e^x))cot(sqrt(e^x))*1/2(e^x)^(-1/2)*d/dx(e^x)#

Here, recall that #d/dx(e^x)=e^x#.

#f'(x)=-csc(sqrt(e^x))cot(sqrt(e^x))*1/2(e^x)^(-1/2)*e^x#

To multiply #(e^x)^(-1/2)*e^x#, add the exponents to get #(e^x)^(1/2)=sqrt(e^x)#.

#f'(x)=(-sqrt(e^x)csc(sqrt(e^x))cot(sqrt(e^x)))/2#

Christopher Agresta · Answer

undefined To differentiate $ f(x) = \csc(\sqrt{e^x}) $ using the chain rule, follow these steps:

1. Identify the outer function as $ \csc(u) $ and the inner function as $ \sqrt{e^x} $.
2. Compute the derivative of the outer function with respect to its argument: $ \frac{d}{du} \csc(u) = -\csc(u) \cot(u) $.
3. Compute the derivative of the inner function with respect to $ x $: $ \frac{d}{dx} \sqrt{e^x} = \frac{1}{2\sqrt{e^x}} e^x = \frac{1}{2\sqrt{e^x}} \cdot e^x $.
4. Apply the chain rule: $ \frac{d}{dx} \csc(\sqrt{e^x}) = -\csc(\sqrt{e^x}) \cot(\sqrt{e^x}) \cdot \frac{1}{2\sqrt{e^x}} \cdot e^x $.

Thus, the derivative of $ f(x) = \csc(\sqrt{e^x}) $ with respect to $ x $ using the chain rule is $ -\frac{e^x}{2\sqrt{e^x}} \csc(\sqrt{e^x}) \cot(\sqrt{e^x}) $.