Chapter 2: Q. 85 (page 235)

Use the quotient rule and the derivative of the sine function to prove that $\frac{d}{d x} (c s x) = - c s c x . c o t x$

Short Answer

Expert verified

The function $\frac{d}{d x} (c s x) = - c s c x . c o t x$ has been proved.

Step by step solution

Step 1. Given Information.

The function:

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

Step 2. Quotient rule.

The quotient rule is:

$(\frac{f}{g})' (x) = \frac{f' (x) g (x) - g' (x) f (x)}{(g {(x))}^{2}}$

Step 3. Obtain the derivative for the given function.

$f' (c s c x) = f' (\frac{1}{\sin x}) = \frac{1' (\sin x) - 1 (\sin x)'}{{(\sin x)}^{2}} = \frac{- \cos x}{\sin^{2} x} = - \frac{1}{\sin x} . \frac{\cos x}{\sin x} = - c s c . c o t x$

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Use the quotient rule and the derivative of the sine function to prove that $\frac{d}{d x} (c s x) = - c s c x . c o t x$

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Quotient rule.

Step 3. Obtain the derivative for the given function.

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Use the quotient rule and the derivative of the sine function to prove that ddx(csx)=-cscx.cotx

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Quotient rule.

Step 3. Obtain the derivative for the given function.

One App. One Place for Learning.

Most popular questions from this chapter

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Use the quotient rule and the derivative of the sine function to prove that $\frac{d}{d x} (c s x) = - c s c x . c o t x$