But that's way too many product rules

Calculus Level 1

d d x ( sin x cos x tan x csc x sec x cot x ) = ? \large \dfrac{d}{dx} ( \sin x \cos x \tan x \csc x \sec x \cot x) = \, ?

0 0 cos x \cos x sin x \sin x tan x \tan x

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2 solutions

Ashish Menon
May 20, 2016

d d x sin x cos x tan x csc x sec x cot x = d d x 1 = 0 \begin{aligned} \dfrac{d}{dx} \sin x \cos x \tan x \csc x \sec x \cot x & = \dfrac{d}{dx} 1\\ & = \boxed{0} \end{aligned}

Applying product rule would prove tedious.

The problem would be much better posed as an MCQ, instead of an integer answer.

Calvin Lin Staff - 5 years ago

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Oh, thank you for your advice I would do it in future. I cant change can you please do that for me? Thanks!

Ashish Menon - 5 years ago

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Sure, done.

Calvin Lin Staff - 5 years ago

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@Calvin Lin Thank you :)

Ashish Menon - 5 years ago

How can we solve d d x sin 1 x cos 1 x tan 1 x csc 1 x sec 1 x cot 1 x \dfrac{d}{dx}\sin^{-1}x\cos^{-1}x\tan^{-1}x\csc^{-1}x\sec^{-1}x\cot^{-1}x ?

Sabhrant Sachan - 5 years ago

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It'll be the same right?

Abhiram Rao - 5 years ago

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Nono Abhiram, sin^-1 is not the same as 1/sin . Same for others ;)

Ashish Menon - 5 years ago

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@Ashish Menon Is that so?

Abhiram Rao - 5 years ago

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@Abhiram Rao Yes cosec and arcsin are different as far as I know.

Ashish Menon - 5 years ago

I dont think that this function is differentiable as it is not continous . it is defined only for x=1,x=-1 as domain of arcsin( x) is [-1,1] and domain of arccosec( x) is[1,infinity)U(-infinity,-1] Please correct me if I am wrong.

AYUSH JAIN - 5 years ago

0 is not one of the given solutions among the multiple choice answers. 1 is, which is incorrect but is accepted as the correct answer.

Ronald Wotzlaw - 5 years ago

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Ah, that's my fault as I was editing the options. Fixed it, thanks!

Calvin Lin Staff - 5 years ago
Eric Ramirez
May 23, 2016

sin is canceled by csc; cos is canceled by sec; tan canceled by cot. 1 is what's left. Derive 1 = 0

That's right. The function expression is equal to 1, except at points where either sin x \sin x or cos x \cos x is equal to zero. Keep in mind that the function does not exist at these points, and its derivative will not exist at these points either.

Pranshu Gaba - 5 years ago

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