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Chapter 3: Problem 29

Verify the following derivative formulas using the Quotient Rule. $$\frac{d}{d x}(\cot x)=-\csc ^{2} x$$

Short Answer

Expert verified
Question: Verify that the derivative of \(\cot x\) is equal to \(-\csc^2 x\) using the Quotient Rule. Answer: To verify the derivative, we first expressed the cotangent function as \(\cot x = \frac{\cos x}{\sin x}\). Then, we applied the Quotient Rule to find the derivative, which resulted in the expression \(h'(x) = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x}\). After simplifying and applying the Pythagorean identity, we found that the derivative of \(\cot x\) is indeed \(-\csc^2 x\).

Step by step solution

01

Express the given function in terms of sine and cosine

Recall that the cotangent function can be written as $$\cot x = \frac{\cos x}{\sin x}$$ So, we will work with this expression.
02

Apply the Quotient Rule

The Quotient Rule states, if a function \(h(x)\) is defined as the quotient of two functions \(u(x)\) and \(v(x)\), then the derivative of \(h(x)\) is given by $$h'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{v^2(x)}$$ Here, \(u(x)=\cos x\) and \(v(x)=\sin x\). Differentiating both with respect to \(x\), we have $$u'(x) = -\sin x \text{ and } v'(x) = \cos x$$ Now, let's find \(h'(x)\) by applying the Quotient Rule. $$h'(x) = \frac{\sin x(-\sin x) - \cos x (\cos x)}{\sin^2 x}$$
03

Simplify the resulting expression

Simplifying the expression above, we get: $$h'(x) = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x}.$$ Recall the Pythagorean identity: $$\sin^2 x + \cos^2 x = 1$$ Then, we can substitute \(1\) in the expression for \(h'(x)\). $$h'(x) = \frac{-1}{\sin^2 x}$$ Finally, recall that \(\csc x = \frac{1}{\sin x}\), and therefore \(\csc^2 x = \frac{1}{\sin^2 x}\). Substituting this result into the expression of \(h'(x)\) gives us: $$h'(x) = -\csc^2 x.$$ We have verified that the derivative of \(\cot x\) is indeed \(-\csc^2 x\), as we were asked to do in the exercise.

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