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Find the derivative of u = tan⁻¹(3*theta2) with respect to theta.

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2 parts
1

dudθ=6θ1+9θ4\frac{du}{d\theta} = \frac{6\theta}{1 + 9\theta^4}

2

−21+x2-\frac{2}{1+x^2}

Step-by-step solution

7 steps

Here are the solutions for the differentiation problems.

Part 1· 3 steps

If u=tan⁡−1(3θ2)u = \tan^{-1}(3\theta^2), find dudx\frac{du}{dx}.

Assumption: The question likely intended to ask for dudθ\frac{du}{d\theta} since the function uu is given in terms of θ\theta. If θ\theta is a function of xx, then dθdx\frac{d\theta}{dx} would be required. We will proceed by finding dudθ\frac{du}{d\theta}.

  1. Identify the function and apply the chain rule for inverse tangent.

    The derivative of tan⁡−1(f(θ))\tan^{-1}(f(\theta)) with respect to θ\theta is f′(θ)1+(f(θ))2\frac{f'(\theta)}{1 + (f(\theta))^2}.
    Here, f(θ)=3θ2f(\theta) = 3\theta^2.

  2. Find the derivative of f(θ)f(\theta) with respect to θ\theta.

    f′(θ)=ddθ(3θ2)=6θf'(\theta) = \frac{d}{d\theta}(3\theta^2) = 6\theta

  3. Substitute f(θ)f(\theta) and f′(θ)f'(\theta) into the derivative formula.

    dudθ=6θ1+(3θ2)2\frac{du}{d\theta} = \frac{6\theta}{1 + (3\theta^2)^2} dudθ=6θ1+9θ4\frac{du}{d\theta} = \frac{6\theta}{1 + 9\theta^4}

    Therefore, assuming the question meant dudθ\frac{du}{d\theta}:

    dudθ=6θ1+9θ4\boxed{\frac{du}{d\theta} = \frac{6\theta}{1 + 9\theta^4}}

Part 2

−21+x2-\frac{2}{1+x^2}

Find ddx[sin⁡−1(1−x21+x2)]\frac{d}{dx}\left[\sin^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right].

  1. Use a trigonometric substitution to simplify the expression inside the inverse sine function.

    Let x=tan⁡θx = \tan\theta. This implies θ=tan⁡−1x\theta = \tan^{-1}x.
    Substitute x=tan⁡θx = \tan\theta into the expression:

    1−x21+x2=1−tan⁡2θ1+tan⁡2θ\frac{1-x^2}{1+x^2} = \frac{1-\tan^2\theta}{1+\tan^2\theta}

    Recall the double angle identity for cosine: cos⁡(2θ)=1−tan⁡2θ1+tan⁡2θ\cos(2\theta) = \frac{1-\tan^2\theta}{1+\tan^2\theta}.
    So, the expression becomes sin⁡−1(cos⁡(2θ))\sin^{-1}(\cos(2\theta)).

  2. Convert cos⁡(2θ)\cos(2\theta) to a sine function using a co-function identity.

    cos⁡(2θ)=sin⁡(π2−2θ)\cos(2\theta) = \sin\left(\frac{\pi}{2} - 2\theta\right)

    Substitute this back into the inverse sine expression:

    sin⁡−1(sin⁡(π2−2θ))=π2−2θ\sin^{-1}\left(\sin\left(\frac{\pi}{2} - 2\theta\right)\right) = \frac{\pi}{2} - 2\theta

  3. Substitute back θ=tan⁡−1x\theta = \tan^{-1}x.

    The expression to differentiate becomes:

    π2−2tan⁡−1x\frac{\pi}{2} - 2\tan^{-1}x

  4. Differentiate the simplified expression with respect to xx.

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

    The derivative of a constant (π2\frac{\pi}{2}) is 0. The derivative of tan⁡−1x\tan^{-1}x is 11+x2\frac{1}{1+x^2}.

    =0−2(11+x2)= 0 - 2\left(\frac{1}{1+x^2}\right) =−21+x2= -\frac{2}{1+x^2} −21+x2\boxed{-\frac{2}{1+x^2}}
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