How do you differentiate #f(x)=cot(1/sqrt(x-3)) # using the chain rule?

Answer 1

Applying a "two-degrees" chain rule.

Just remembering the concept: #(dy)/(dx)=(dy)/(du)(du)/(dv)(dv)/(dx)#

Here, we can name #u=(v)^(-1/2)# and #v=x-3#
Also, we must remember that the rule to derivate #cota# is
#cota=-a'csc^2a#
#(dy)/(dx)=-u'csc^2u*(-1/(2v^(3/2)))*(1)#
#(dy)/(dx)=-(-1/(2v^(3/2)))csc^2(v^(-1/2))*(-1/(2(x-3)^(3/2)))#
#(dy)/(dx)=(csc^2(x-3)^(-1/2))/(2(x-3)^(3/2))(-1/(2(x-3)^(3/2)))#
#(dy)/(dx)=-(csc^2(1/(x-3)^(1/2)))/(4(x-3)^3)#
#(dy)/(dx)=-(csc^2(1/(sqrt(x-3))))/(4(x-3)^3)#
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Answer 2

To differentiate ( f(x) = \cot\left(\frac{1}{\sqrt{x-3}}\right) ) using the chain rule, follow these steps:

  1. Let ( u = \frac{1}{\sqrt{x-3}} ).
  2. Find the derivative of ( u ) with respect to ( x ): ( \frac{du}{dx} = -\frac{1}{2(x-3)^{3/2}} ).
  3. Apply the chain rule: ( \frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx} ).
  4. Now, differentiate ( f(u) = \cot(u) ) with respect to ( u ): ( \frac{df}{du} = -\csc^2(u) ).
  5. Substitute ( u = \frac{1}{\sqrt{x-3}} ) and ( \frac{du}{dx} = -\frac{1}{2(x-3)^{3/2}} ) into the chain rule formula.
  6. ( \frac{df}{dx} = -\csc^2\left(\frac{1}{\sqrt{x-3}}\right) \cdot \left(-\frac{1}{2(x-3)^{3/2}}\right) ).
  7. Simplify the expression: ( \frac{df}{dx} = \frac{1}{2(x-3)^{3/2}\sin^2\left(\frac{1}{\sqrt{x-3}}\right)} ).
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Answer from HIX Tutor

When evaluating a one-sided 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 one-sided 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 one-sided 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 one-sided 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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