Resolved: "Prove that ddx(csc⁡x)=csc⁡xcot⁡x"

FobelloE

Answered question

2021-10-30

Prove that

\frac{d}{d x} (\csc x) = \csc x \cot x

tabuordg

Skilled2021-10-31Added 99 answers

\frac{d (\csc x)}{d x} = \frac{d}{d x} [\frac{1}{\sin x}]

The Quotient Rule for differentiation

\frac{d}{d x} [\frac{f (x)}{g (x)}] = \frac{f^{'} (x) \cdot g (x) - f (x) \cdot g^{'} (x)}{{[g (x)]}^{2}}

= \frac{{(1)}^{'} \cdot (\sin x) - 1 \cdot {(\sin x)}^{'}}{{(\sin x)}^{2}}

Remember that:
1) The derivative of

\sin x

\cos x

2) The derivative of a constant is 0

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

= \frac{0 - \cos x}{\sin^{2} x}

= \frac{- 1}{\sin x} \cdot \frac{\cos x}{\sin x}

Note that:

\frac{1}{\sin θ} = \csc θ

and

\frac{\cos θ}{\sin θ} = \cot θ

= - \csc x \cot x

Result: Use the quotient rule for differentiation

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