Bifunctional Derivative: Trigonometric and Logarithmic

Calculus Level 2

Let f ( x ) = ln ⁡ ( sin ⁡ ( x ) cot ⁡ ( x ) ) f(x) = \ln( \sin(x) \cot(x)) , find f ′ ( π 4 ) f'\left(\dfrac{\pi}{4}\right) .

1 1 ∞ \infty − 1 -1 − ∞ -\infty 0 0 Undefined at the given x x value

David Hontz by David Hontz , 26, USA

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1 solution

David Hontz
May 13, 2016

First: c o t ( x ) = c o s ( x ) s i n ( x ) cot(x) = \frac{cos(x)}{sin(x)} ; therefore, s i n ( x ) c o t ( x ) = s i n ( x ) c o s ( x ) s i n ( x ) = c o s ( x ) sin(x)cot(x) = \frac{sin(x) cos(x)}{sin(x)} = cos(x)

Now: f ( x ) = l n [ c o s ( x ) ] f(x) = ln[cos(x)]

f ′ ( x ) = 1 c o s ( x ) − s i n ( x ) 1 = − s i n ( x ) c o s ( x ) = − t a n ( x ) f'(x) = \frac{1}{cos(x)} \frac{-sin(x)}{1} = \frac{-sin(x)}{cos(x)} = -tan(x)

Finally, plug in π 4 \frac{π}{4} into − t a n ( x ) -tan(x)

f ′ ( π 4 ) = − 1 f'(\frac{π}{4}) = -1

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