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Find the derivative of ^-1x + ^-1x, x, and ^-1sqrt(1+x2) + ^-1sqrt(1+x2) with respect to x.

Mathematics

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Question 2:
We need to find the derivative of sec⁡−1x+csc⁡−1x\sec^{-1}x + \csc^{-1}x with respect to xx.

Step 1: Recall the inverse trigonometric identity sec⁡−1x+csc⁡−1x=π2\sec^{-1}x + \csc^{-1}x = \frac{\pi}{2}.

Step 2: Substitute the identity into the expression.

ddx(sec⁡−1x+csc⁡−1x)=ddx(π2)\frac{d}{dx}(\sec^{-1}x + \csc^{-1}x) = \frac{d}{dx}\left(\frac{\pi}{2}\right)

Step 3: Differentiate the constant.

ddx(π2)=0\frac{d}{dx}\left(\frac{\pi}{2}\right) = 0

The correct option is (b).
The final answer is 0\boxed{\text{0}}.

Question 3:
We need to find the derivative of sec⁡x\sec x with respect to xx.

Step 1: Recall the standard derivative of sec⁡x\sec x.

ddx(sec⁡x)=sec⁡xtan⁡x\frac{d}{dx}(\sec x) = \sec x \tan x

The correct option is (a).
The final answer is sec⁡xtan⁡x\boxed{\sec x \tan x}.

Question 4:
We need to find the derivative of tan⁡−11+x2+cot⁡−11+x2\tan^{-1}\sqrt{1+x^2} + \cot^{-1}\sqrt{1+x^2} with respect to xx.

Step 1: Recall the inverse trigonometric identity tan⁡−1u+cot⁡−1u=π2\tan^{-1}u + \cot^{-1}u = \frac{\pi}{2}. Here, u=1+x2u = \sqrt{1+x^2}.

Step 2: Substitute the identity into the expression.

ddx(tan⁡−11+x2+cot⁡−11+x2)=ddx(π2)\frac{d}{dx}(\tan^{-1}\sqrt{1+x^2} + \cot^{-1}\sqrt{1+x^2}) = \frac{d}{dx}\left(\frac{\pi}{2}\right)

Step 3: Differentiate the constant.

ddx(π2)=0\frac{d}{dx}\left(\frac{\pi}{2}\right) = 0

The correct option is (d).
The final answer is 0\boxed{\text{0}}.

Question 5:
We need to find dd(3x)(2x)\frac{d}{d(3x)}(2^x). This is a derivative with respect to 3x3x.

Step 1: Let y=2xy = 2^x and u=3xu = 3x. We want to find dydu\frac{dy}{du}.

Step 2: Find dydx\frac{dy}{dx} and dudx\frac{du}{dx}.

dydx=ddx(2x)=2xln⁡2\frac{dy}{dx} = \frac{d}{dx}(2^x) = 2^x \ln 2 dudx=ddx(3x)=3\frac{du}{dx} = \frac{d}{dx}(3x) = 3

Step 3: Use the chain rule dydu=dy/dxdu/dx\frac{dy}{du} = \frac{dy/dx}{du/dx}.

dydu=2xln⁡23\frac{dy}{du} = \frac{2^x \ln 2}{3}

The result is 132xln⁡2\frac{1}{3} 2^x \ln 2, which does not match options (a), (b), or (c).
The correct option is (d).
The final answer is 132xln⁡2\boxed{\frac{1}{3} 2^x \ln 2}.

Question 6:
Given y=tan⁡−1[cos⁡x1+sin⁡x]y = \tan^{-1}\left[\frac{\cos x}{1+\sin x}\right], find dydx\frac{dy}{dx}.

Step 1: Simplify the expression inside the tan⁡−1\tan^{-1} function.

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