How do you differentiate #f(x)=e^cot(sqrt(x)) # using the chain rule?
# f'(x) = (csc^2(sqrt(x)) * e^(cot sqrt(x))) / (2sqrt(x)) #
f you are studying maths, then you should learn the Chain Rule for Differentiation, and practice how to use it:
{ ("Let",u=sqrt(x)=x^(1/2), => , (du)/dx=1/2x^(1/2)=1/(2sqrt(x))),
("And",v=cot sqrt(x)=cotu, => , (dv)/(du)=csc^2u), ("Then",y=e^(cot sqrt(x))=e^v, =>, dy/(dv)=e^v ) :}#
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To differentiate ( f(x) = e^{\cot(\sqrt{x})} ) using the chain rule:

Identify the outer function ( u ) and the inner function ( v ). ( u = e^v ) and ( v = \cot(\sqrt{x}) ).

Differentiate the outer function with respect to the inner function: ( \frac{du}{dv} = e^v ).

Differentiate the inner function with respect to the variable of differentiation (in this case, ( x )): ( \frac{dv}{dx} = \frac{d}{dx}[\cot(\sqrt{x})] ).

Apply the chain rule: ( \frac{du}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} ).

Substitute the derivatives calculated in steps 2 and 3 into step 4 and simplify to obtain the final result.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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