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How do you differentiate #f(x) = 1/sqrt(arctan(e^(x-1)) # using the chain rule?

Answer 1

Ezra Adams

#f'(x) = frac{-e^(x-1)}{2((e^(x-1))^2+1)sqrt((arctan(e^(x-1)))^3)}#

Let #f(x) = [arctan(e^(x-1))]^(-1/2)#

Use the chain rule, which says that #color(blue)(d/dx(f(g(x))) = f'(g(x))*g'(x))#:

#f'(x) = (-1/2)(arctan(e^(x-1)))^(-3/2)*(frac{1}{(e^(x-1))^2+1})*e^(x-1)#

Now we simplify to get: #f'(x) = frac{-e^(x-1)}{2((e^(x-1))^2+1)sqrt((arctan(e^(x-1)))^3)}#

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Answer 2

Samantha Adkison

To differentiate ( f(x) = \frac{1}{\sqrt{\arctan(e^{x-1})}} ) using the chain rule, we first identify the outer function ( g(x) = \sqrt{x} ) and the inner function ( h(x) = \arctan(e^{x-1}) ). Then we differentiate the outer function with respect to the inner function, followed by differentiating the inner function with respect to ( x ). The derivative is:

[ f'(x) = -\frac{1}{2(\arctan(e^{x-1}))^{3/2}} \cdot \frac{1}{1 + (e^{x-1})^2} \cdot e^{x-1} ]

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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.