How do you differentiate # (sec(πx))/(tan(πx))#?

Answer 1

#= -pi csc pi x cot pi x#

#d/dx (sec(πx))/(tan(πx))#
#=d/dx (cos pi x)/(cos pi x sin pi x)#
#=d/dx ( sin pi x)^(-1) [= d/dx csc pi x]#
by power and chain rules #= -( sin pi x)^(-2) * cos pi x * pi#
#= -(pi cos pi x)/( sin^2 pi x)#
#= -pi csc pi x cot pi x#
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Answer 2

To differentiate (\frac{{\sec(\pi x)}}{{\tan(\pi x)}}), we can use the quotient rule, which states that if (f(x) = \frac{{g(x)}}{{h(x)}}), then (f'(x) = \frac{{g'(x)h(x) - g(x)h'(x)}}{{[h(x)]^2}}).

Given (g(x) = \sec(\pi x)) and (h(x) = \tan(\pi x)), we differentiate each function separately to find (g'(x)) and (h'(x)).

(g'(x) = \frac{{d}}{{dx}}(\sec(\pi x)) = \pi\sec(\pi x)\tan(\pi x))

(h'(x) = \frac{{d}}{{dx}}(\tan(\pi x)) = \pi\sec^2(\pi x))

Now, apply the quotient rule:

(f'(x) = \frac{{\pi\sec(\pi x)\tan(\pi x)\cdot\tan(\pi x) - \sec(\pi x)\cdot\pi\sec^2(\pi x)}}{{[\tan(\pi x)]^2}})

(f'(x) = \frac{{\pi\sec(\pi x)\tan^2(\pi x) - \pi\sec^3(\pi x)}}{{\tan^2(\pi x)}})

(f'(x) = \frac{{\pi\sec(\pi x)[\tan^2(\pi x) - \sec^2(\pi x)]}}{{\tan^2(\pi x)}})

(f'(x) = \frac{{\pi\sec(\pi x)[\tan^2(\pi x) - \sec^2(\pi x)]}}{{\tan^2(\pi x)}})

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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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