Chapter 2: Q. 86 (page 235)

Use the quotient rule and the derivatives of the sine and cosine functions to prove that $\frac{d}{d x} (c o t x) = - c s c^{2} x$ .

Short Answer

Expert verified

We proved $\frac{d}{d x} (c o t x) = - c s c^{2} x$ .

Step by step solution

Step 1. Given Information

We are given $c o t x$ and using the quotient rule and the derivatives of the sine and cosine functions we need to prove that $\frac{d}{d x} (c o t x) = - c s c^{2} x$ .

Step 2. Proving the expression

$\frac{d}{d x} (c o t x) = \frac{d}{d x} (\frac{\cos x}{\sin x})$

Using the quotient rule we get,

$= \frac{(\frac{d}{d x} (\cos x)) \sin x - \cos x (\frac{d}{d x} (\sin x))}{\sin^{2} x} = \frac{- \sin x \sin x - \cos x \cos x}{\sin^{2} x} = \frac{- (\sin^{2} x + \cos^{2} x)}{\sin^{2} x} = \frac{- 1}{\sin^{2} x} = - c s c^{2} x$

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Use the quotient rule and the derivatives of the sine and cosine functions to prove that $\frac{d}{d x} (c o t x) = - c s c^{2} x$ .

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Proving the expression

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Use the quotient rule and the derivatives of the sine and cosine functions to prove thatddxcotx=-csc2x.

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Proving the expression

One App. One Place for Learning.

Most popular questions from this chapter

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Use the quotient rule and the derivatives of the sine and cosine functions to prove that $\frac{d}{d x} (c o t x) = - c s c^{2} x$ .