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What is the derivative of #f(x)=1/sinx#?

Answer 1

Elijah Abram

Since #f(x)=1/(sin(x))=csc(x)#, the answer can be written down from memorization that #f'(x)=-csc(x)cot(x)#.

Alternatively, the Quotient Rule can be used: #f'(x)=\frac{sin(x)*0-1*cos(x)}{sin^{2}(x)}=-\frac{cos(x)}{sin^{2}(x)}#

#=-\frac{1}{sin(x)}*\frac{cos(x)}{sin(x)}=-csc(x)cot(x)#

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Answer 2

Emma Aldridge

To find the derivative of ( f(x) = \frac{1}{\sin(x)} ), we can use the quotient rule. The quotient rule 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} ). Applying this rule:

Let ( g(x) = 1 ) and ( h(x) = \sin(x) ).

( g'(x) = 0 ) (derivative of a constant is zero).

( h'(x) = \cos(x) ) (derivative of sin(x) is cos(x)).

Now, applying the quotient rule:

( f'(x) = \frac{0 \cdot \sin(x) - 1 \cdot \cos(x)}{[\sin(x)]^2} )

( f'(x) = \frac{-\cos(x)}{\sin^2(x)} )

So, the derivative of ( f(x) = \frac{1}{\sin(x)} ) is ( f'(x) = -\frac{\cos(x)}{\sin^2(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.