Show that the derivative of cot(x) is -csc2(x) using the quotient rule.

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To show that d/dx(cot x) = −csc² x using the quotient rule, we start with cot x = (cos x)/(sin x). Let u = cos x and v = sin x. Step 1: Find the derivatives of u and v. du/dx = d/dx(cos x) = −sin x; dv/dx = d/dx(sin x) = cos x; Step 2: Apply the quotient rule, which states that if y = u/v, then dy/dx = (v du/dx − u dv/dx)/(v²). d/dx(cot x) = ((sin x)(−sin x) − (cos x)(cos x))/((sin x)²) Step 3: Simplify the numerator. d/dx(cot x) = (−sin² x − cos² x)/(sin² x) Step 4: Factor out −1 from the numerator. d/dx(cot x) = (−(sin² x + cos² x))/(sin² x); Step 5: Use the Pythagorean identity sin² x + cos² x = 1. d/dx(cot x) = (−1)/(sin² x); Step 6: Express the result in terms of csc x, where csc x = (1)/(sin x). d/dx(cot x) = −((1)/(sin x))² = −csc² x Thus, we have shown that: d/dx(cot x) = −csc² x

Show that the derivative of cot(x) is -csc2(x) using the quotient rule.

Answer

Step-by-step solution

Find the derivatives of $u$ and $v$ .

Apply the quotient rule, which states that if $y = \frac{u}{v}$ , then $\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$ .

Simplify the numerator.

Factor out $-1$ from the numerator.

Use the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ .

Express the result in terms of $\csc x$ , where $\csc x = \frac{1}{\sin x}$ .

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Show that the derivative of cot(x) is -csc2(x) using the quotient rule.

Answer

Step-by-step solution

Find the derivatives of uuu and vvv.

Apply the quotient rule, which states that if y=uvy = \frac{u}{v}y=vu​, then dydx=vdudx−udvdxv2\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}dxdy​=v2vdxdu​−udxdv​​.

Simplify the numerator.

Factor out −1-1−1 from the numerator.

Use the Pythagorean identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1sin2x+cos2x=1.

Express the result in terms of csc⁡x\csc xcscx, where csc⁡x=1sin⁡x\csc x = \frac{1}{\sin x}cscx=sinx1​.

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More Mathematics Questions

Find the derivatives of $u$ and $v$ .

Apply the quotient rule, which states that if $y = \frac{u}{v}$ , then $\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$ .

Factor out $-1$ from the numerator.

Use the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ .

Express the result in terms of $\csc x$ , where $\csc x = \frac{1}{\sin x}$ .