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Identify u and v for the quotient rule (d)/(dx)((u)/(v)) = (u'v - uv')/(v²).

Mathematics

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Here are the derivatives for the remaining functions:

2. f(x)=ln⁡(x)x2sin⁡(x)f(x) = \frac{\ln(x)}{x^2 \sin(x)}

  1. Identify uu and vv for the quotient rule ddx(uv)=u′v−uv′v2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}.

    Let u=ln⁡(x)u = \ln(x) and v=x2sin⁡(x)v = x^2 \sin(x).

  2. Find the derivatives of uu and vv.

    For uu: u′=1xu' = \frac{1}{x}.
    For vv, use the product rule: v′=ddx(x2)sin⁡(x)+x2ddx(sin⁡(x))=2xsin⁡(x)+x2cos⁡(x)v' = \frac{d}{dx}(x^2)\sin(x) + x^2\frac{d}{dx}(\sin(x)) = 2x\sin(x) + x^2\cos(x).

  3. Apply the quotient rule.

    f′(x)=(1x)(x2sin⁡(x))−(ln⁡(x))(2xsin⁡(x)+x2cos⁡(x))(x2sin⁡(x))2f'(x) = \frac{\left(\frac{1}{x}\right)(x^2 \sin(x)) - (\ln(x))(2x\sin(x) + x^2\cos(x))}{(x^2 \sin(x))^2}
  4. Simplify the expression.

    f′(x)=xsin⁡(x)−2xln⁡(x)sin⁡(x)−x2ln⁡(x)cos⁡(x)x4sin⁡2(x)f'(x) = \frac{x\sin(x) - 2x\ln(x)\sin(x) - x^2\ln(x)\cos(x)}{x^4 \sin^2(x)}

    Factor out xx from the numerator.

    f′(x)=x(sin⁡(x)−2ln⁡(x)sin⁡(x)−xln⁡(x)cos⁡(x))x4sin⁡2(x)f'(x) = \frac{x(\sin(x) - 2\ln(x)\sin(x) - x\ln(x)\cos(x))}{x^4 \sin^2(x)}

    Cancel one xx from numerator and denominator.

    f′(x)=sin⁡(x)−2ln⁡(x)sin⁡(x)−xln⁡(x)cos⁡(x)x3sin⁡2(x)f'(x) = \boxed{\frac{\sin(x) - 2\ln(x)\sin(x) - x\ln(x)\cos(x)}{x^3 \sin^2(x)}}

    4. y=tan⁡(x)e3xy = \frac{\tan(x)}{e^{3x}}

  5. Identify uu and vv for the quotient rule.

    Let u=tan⁡(x)u = \tan(x) and v=e3xv = e^{3x}.

  6. Find the derivatives of uu and vv.

    For uu: u′=sec⁡2(x)u' = \sec^2(x).
    For vv, use the chain rule: v′=e3x⋅ddx(3x)=3e3xv' = e^{3x} \cdot \frac{d}{dx}(3x) = 3e^{3x}.

  7. Apply the quotient rule.

    dydx=(sec⁡2(x))(e3x)−(tan⁡(x))(3e3x)(e3x)2\frac{dy}{dx} = \frac{(\sec^2(x))(e^{3x}) - (\tan(x))(3e^{3x})}{(e^{3x})^2}
  8. Simplify the expression.

    dydx=e3x(sec⁡2(x)−3tan⁡(x))e6x\frac{dy}{dx} = \frac{e^{3x}(\sec^2(x) - 3\tan(x))}{e^{6x}}

    Cancel e3xe^{3x} from numerator and denominator.

    dydx=sec⁡2(x)−3tan⁡(x)e3x\frac{dy}{dx} = \boxed{\frac{\sec^2(x) - 3\tan(x)}{e^{3x}}}

    6. f(x)=2xx2f(x) = \frac{2x}{x^2}

  9. Simplify the function first.

    f(x)=2x=2x−1f(x) = \frac{2}{x} = 2x^{-1}
  10. Differentiate the simplified function using the power rule.

    f′(x)=2⋅(−1)x−1−1f'(x) = 2 \cdot (-1)x^{-1-1} f′(x)=−2x−2f'(x) = -2x^{-2} f′(x)=−2x2f'(x) = \boxed{-\frac{2}{x^2}}

    8. y=ln⁡(x2+1)xy = \frac{\ln(x^2 + 1)}{\sqrt{x}}

  11. Identify uu and vv for the quotient rule.

    Let u=ln⁡(x2+1)u = \ln(x^2 + 1) and v=x=x1/2v = \sqrt{x} = x^{1/2}.

  12. Find the derivatives of uu and vv.

    For uu, use the chain rule: u′=1x2+1⋅ddx(x2+1)=2xx2+1u' = \frac{1}{x^2 + 1} \cdot \frac{d}{dx}(x^2 + 1) = \frac{2x}{x^2 + 1}.
    For vv: v′=12x−1/2=12xv' = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}.

  13. Apply the quotient rule.

    dydx=(2xx2+1)(x)−(ln⁡(x2+1))(12x)(x)2\frac{dy}{dx} = \frac{\left(\frac{2x}{x^2 + 1}\right)(\sqrt{x}) - (\ln(x^2 + 1))\left(\frac{1}{2\sqrt{x}}\right)}{(\sqrt{x})^2}
  14. Simplify the expression.

    Multiply the numerator and denominator by 2x(x2+1)2\sqrt{x}(x^2+1) to clear fractions in the numerator.

    dydx=2xxx2+1−ln⁡(x2+1)2xx⋅2x(x2+1)2x(x2+1)\frac{dy}{dx} = \frac{\frac{2x\sqrt{x}}{x^2 + 1} - \frac{\ln(x^2 + 1)}{2\sqrt{x}}}{x} \cdot \frac{2\sqrt{x}(x^2+1)}{2\sqrt{x}(x^2+1)} dydx=2xx(2x)−ln⁡(x2+1)(x2+1)x⋅2x(x2+1)\frac{dy}{dx} = \frac{2x\sqrt{x}(2\sqrt{x}) - \ln(x^2 + 1)(x^2+1)}{x \cdot 2\sqrt{x}(x^2+1)} dydx=4x2−(x2+1)ln⁡(x2+1)2x3/2(x2+1)\frac{dy}{dx} = \frac{4x^2 - (x^2+1)\ln(x^2 + 1)}{2x^{3/2}(x^2+1)} dydx=4x2−(x2+1)ln⁡(x2+1)2xx(x2+1)\frac{dy}{dx} = \boxed{\frac{4x^2 - (x^2+1)\ln(x^2 + 1)}{2x\sqrt{x}(x^2+1)}}

    10. f(x)=exln⁡xx2+1f(x) = \frac{e^x \ln x}{x^2 + 1}

  15. Identify uu and vv for the quotient rule.

    Let u=exln⁡xu = e^x \ln x and v=x2+1v = x^2 + 1.

  16. Find the derivatives of uu and vv.

    For uu, use the product rule: u′=ddx(ex)ln⁡x+exddx(ln⁡x)=exln⁡x+ex⋅1x=ex(ln⁡x+1x)u' = \frac{d}{dx}(e^x)\ln x + e^x\frac{d}{dx}(\ln x) = e^x \ln x + e^x \cdot \frac{1}{x} = e^x\left(\ln x + \frac{1}{x}\right).
    For vv: v′=2xv' = 2x.

  17. Apply the quotient rule.

    f′(x)=(ex(ln⁡x+1x))(x2+1)−(exln⁡x)(2x)(x2+1)2f'(x) = \frac{\left(e^x\left(\ln x + \frac{1}{x}\right)\right)(x^2 + 1) - (e^x \ln x)(2x)}{(x^2 + 1)^2}
  18. Simplify the expression.

    Factor out exe^x from the numerator.

    f′(x)=ex[(ln⁡x+1x)(x2+1)−2xln⁡x](x2+1)2f'(x) = \frac{e^x\left[\left(\ln x + \frac{1}{x}\right)(x^2 + 1) - 2x \ln x\right]}{(x^2 + 1)^2}

    Expand the term in the brackets.

    f′(x)=ex[x2ln⁡x+ln⁡x+x+1x−2xln⁡x](x2+1)2f'(x) = \frac{e^x\left[x^2 \ln x + \ln x + x + \frac{1}{x} - 2x \ln x\right]}{(x^2 + 1)^2}

    Combine like terms.

    f′(x)=ex[x2ln⁡x−2xln⁡x+ln⁡x+x+1x](x2+1)2f'(x) = \boxed{\frac{e^x\left[x^2 \ln x - 2x \ln x + \ln x + x + \frac{1}{x}\right]}{(x^2 + 1)^2}}
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