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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 the derivative of the cotangent function using the Quotient Rule. Answer: Using the Quotient Rule, we derived and simplified the derivative of the cotangent function and found that $$\frac{d}{d x}(\cot x)=-\csc^{2}x$$. This confirms the given derivative formula for the cotangent function.

Step by step solution

01

Recall the Cotangent Function

The cotangent function can be written as the quotient of the cosine function and the sine function: $$\cot x=\frac{\cos x}{\sin x}$$. We will now use the Quotient Rule to find the derivative of the cotangent function. Step 2: Apply the Quotient Rule
02

Apply the Quotient Rule

The Quotient Rule states that for two functions u(x) and v(x), the derivative of their quotient is given by: $$\frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$$ In our case, we have u(x) = cos(x) and v(x) = sin(x). First, we find the derivatives of u(x) and v(x): $$\frac{d}{dx}(\cos x) = -\sin x$$ and $$\frac{d}{dx}(\sin x) = \cos x$$ Now, apply the Quotient Rule: $$\frac{d}{d x}(\cot x)=\frac{d}{d x}\left(\frac{\cos x}{\sin x}\right)=\frac{\sin x\left(-\sin x\right)-\cos x\left(\cos x\right)}{(\sin x)^{2}}$$ Step 3: Simplify the Result
03

Simplify the Derivative

Now, we simplify the resulting derivative: $$\frac{d}{d x}(\cot x)=\frac{-\sin^{2}x - \cos^{2}x}{\sin^{2}x}$$ Recall that the Pythagorean trigonometric identity states that: $$\sin^2 x + \cos^2 x = 1$$, which implies$$-\sin^2 x - \cos^2 x = -1$$ So, we can write: $$\frac{d}{d x}(\cot x)=\frac{-1}{\sin^{2}x}$$ Step 4: Express the Result in Terms of the Cosecant Function
04

Use the Cosecant Function

Finally, we can express the result of the derivative as a function of the cosecant function, which is the reciprocal of the sine function: $$\frac{d}{d x}(\cot x)=-\csc^{2}x$$ Therefore, we have verified the given derivative formula using the Quotient Rule.

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Most popular questions from this chapter

Special Quotient Rule In general, the derivative of a quotient is not the quotient of the derivatives. Find nonconstant functions \(f\) and \(g\) such that the derivative of \(f / g\) equals \(f^{\prime} / g^{\prime}\).

Graphing \(f\) and \(f^{\prime}\) a. Graph \(f\) with a graphing utility. b. Compute and graph \(f^{\prime}\) c. Verify that the zeros of \(f^{\prime}\) correspond to points at which \(f\) has \(a\) horizontal tangent line. $$f(x)=(x-1) \sin ^{-1} x \text { on }[-1,1]$$

Use the following table to find the given derivatives. $$\begin{array}{llllll} x & 1 & 2 & 3 & 4 & 5 \\ \hline f(x) & 5 & 4 & 3 & 2 & 1 \\ f^{\prime}(x) & 3 & 5 & 2 & 1 & 4 \\ g(x) & 4 & 2 & 5 & 3 & 1 \\ g^{\prime}(x) & 2 & 4 & 3 & 1 & 5 \end{array}$$ $$\left.\frac{d}{d x}\left[\frac{x f(x)}{g(x)}\right]\right|_{x=4}$$

Volume of a torus The volume of a torus (doughnut or bagel) with an inner radius of \(a\) and an outer radius of \(b\) is \(V=\pi^{2}(b+a)(b-a)^{2} / 4\) a. Find \(d b / d a\) for a torus with a volume of \(64 \pi^{2}\). b. Evaluate this derivative when \(a=6\) and \(b=10\)

Prove the following identities and give the values of \(x\) for which they are true. $$\cos \left(\sin ^{-1} x\right)=\sqrt{1-x^{2}}$$
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