/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 Verify the following derivative ... [FREE SOLUTION] | 91Ó°ÊÓ

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Verify the following derivative formulas using the Quotient Rule. $$\frac{d}{d x}(\csc x)=-\csc x \cot x$$

Short Answer

Expert verified
Question: Use the Quotient Rule to verify the derivative formula for \(\csc x\), which is \(\frac{d}{d x}(\csc x) = -\csc x \cot x\). Answer: Follow the steps below to verify the derivative formula for \(\csc x\). Step 1: Write csc(x) as a fraction \(\csc x = \frac{1}{\sin x}\) Step 2: Apply the Quotient Rule \(\frac{d}{d x}(\frac{1}{\sin x}) = \frac{-\cos x}{\sin^2 x}\) Step 3: Simplify the expression \(\frac{d}{d x}(\frac{1}{\sin x}) = \frac{-\cos x}{\sin^2 x}\) Step 4: Use trigonometric identities to rewrite the expression \(\frac{-\cos x}{\sin^2 x} = -\csc x \cot x\) Step 5: Conclusion \(\frac{d}{d x}(\csc x)=-\csc x \cot x\) The derivative formula for \(\csc x\) is verified using the Quotient Rule.

Step by step solution

01

Write csc(x) as a fraction

We can rewrite \(\csc x\) as \(\frac{1}{\sin x}\): $$\csc x = \frac{1}{\sin x}$$ Now, we can use the Quotient Rule to find the derivative of \(\csc x\) with respect to \(x\).
02

Apply the Quotient Rule

Using the Quotient Rule, let's find the derivative of \(\frac{1}{\sin x}\) with respect to \(x\). Let \(u(x) = 1\) and \(v(x) = \sin x\). First, find the derivatives of \(u(x)\) and \(v(x)\) with respect to x: $$u'(x)=\frac{d}{d x}(1) = 0 $$ $$v'(x)=\frac{d}{d x}(\sin x) = \cos x$$ Now, apply the Quotient Rule: $$\frac{d}{d x}(\frac{1}{\sin x}) = \frac{(0)(\sin x) - (1)(\cos x)}{[\sin x]^2}$$
03

Simplify the expression

Now, simplify the expression: $$\frac{d}{d x}(\frac{1}{\sin x}) = \frac{-\cos x}{\sin^2 x}$$
04

Use trigonometric identities to rewrite the expression

We can rewrite the expression in terms of \(\csc x\) and \(\cot x\) using trigonometric identities. Recall that \(\csc x = \frac{1}{\sin x}\) and \(\cot x = \frac{\cos x}{\sin x}\): $$\frac{-\cos x}{\sin^2 x} = -\frac{1}{\sin x}\frac{\cos x}{\sin x} = -\csc x \cot x$$
05

Conclusion

We found that the derivative of \(\csc x\) with respect to \(x\) is indeed equal to \(-\csc x \cot x\): $$\frac{d}{d x}(\csc x)=-\csc x \cot x$$ The derivative formula has been verified using the Quotient Rule.

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