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Problem 6: Find the derivative of (x+y) = y² x.

Mathematics

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: Find the derivative of sin⁡(x+y)=y2cos⁡x\sin(x+y) = y^2 \cos x.

We use implicit differentiation with respect to xx.

  1. Differentiate both sides of the equation.

    ddx(sin⁡(x+y))=ddx(y2cos⁡x)\frac{d}{dx}(\sin(x+y)) = \frac{d}{dx}(y^2 \cos x)

  2. Apply the chain rule on the left side and the product rule on the right side.

    cos⁡(x+y)⋅ddx(x+y)=ddx(y2)⋅cos⁡x+y2⋅ddx(cos⁡x)\cos(x+y) \cdot \frac{d}{dx}(x+y) = \frac{d}{dx}(y^2) \cdot \cos x + y^2 \cdot \frac{d}{dx}(\cos x) cos⁡(x+y)(1+dydx)=2ydydxcos⁡x+y2(−sin⁡x)\cos(x+y) \left(1 + \frac{dy}{dx}\right) = 2y \frac{dy}{dx} \cos x + y^2 (-\sin x) cos⁡(x+y)+cos⁡(x+y)dydx=2ycos⁡xdydx−y2sin⁡x\cos(x+y) + \cos(x+y) \frac{dy}{dx} = 2y \cos x \frac{dy}{dx} - y^2 \sin x

  3. Group terms containing dydx\frac{dy}{dx} on one side and other terms on the other side.

    cos⁡(x+y)dydx−2ycos⁡xdydx=−y2sin⁡x−cos⁡(x+y)\cos(x+y) \frac{dy}{dx} - 2y \cos x \frac{dy}{dx} = -y^2 \sin x - \cos(x+y)

  4. Factor out dydx\frac{dy}{dx} and solve for it.

    dydx(cos⁡(x+y)−2ycos⁡x)=−(y2sin⁡x+cos⁡(x+y))\frac{dy}{dx} (\cos(x+y) - 2y \cos x) = -(y^2 \sin x + \cos(x+y)) dydx=−(y2sin⁡x+cos⁡(x+y))cos⁡(x+y)−2ycos⁡x=y2sin⁡x+cos⁡(x+y)2ycos⁡x−cos⁡(x+y)\boxed{\frac{dy}{dx} = \frac{-(y^2 \sin x + \cos(x+y))}{\cos(x+y) - 2y \cos x} = \frac{y^2 \sin x + \cos(x+y)}{2y \cos x - \cos(x+y)}}

    : Find the derivative of f(x)=arctan⁡(2x1−x2)f(x) = \arctan\left(\frac{2x}{1-x^2}\right).

    We can use a trigonometric substitution to simplify the function before differentiating.

  5. Let x=tan⁡θx = \tan\theta.

    Then θ=arctan⁡x\theta = \arctan x.

    f(x)=arctan⁡(2tan⁡θ1−tan⁡2θ)f(x) = \arctan\left(\frac{2\tan\theta}{1-\tan^2\theta}\right)

  6. Use the double angle identity tan⁡(2θ)=2tan⁡θ1−tan⁡2θ\tan(2\theta) = \frac{2\tan\theta}{1-\tan^2\theta}.

    f(x)=arctan⁡(tan⁡(2θ))f(x) = \arctan(\tan(2\theta)) f(x)=2θf(x) = 2\theta

  7. Substitute θ=arctan⁡x\theta = \arctan x back into the function.

    f(x)=2arctan⁡xf(x) = 2 \arctan x

  8. Differentiate f(x)f(x) with respect to xx.

    f′(x)=ddx(2arctan⁡x)f'(x) = \frac{d}{dx}(2 \arctan x) f′(x)=2⋅11+x2f'(x) = 2 \cdot \frac{1}{1+x^2} f′(x)=21+x2\boxed{f'(x) = \frac{2}{1+x^2}}

    : Find the derivative of y=ln⁡(sinh⁡x+cosh⁡x)y = \ln(\sinh x + \cosh x).

  9. Simplify the expression inside the logarithm.

    Recall that sinh⁡x=ex−e−x2\sinh x = \frac{e^x - e^{-x}}{2} and cosh⁡x=ex+e−x2\cosh x = \frac{e^x + e^{-x}}{2}.

    sinh⁡x+cosh⁡x=ex−e−x2+ex+e−x2=ex−e−x+ex+e−x2=2ex2=ex\sinh x + \cosh x = \frac{e^x - e^{-x}}{2} + \frac{e^x + e^{-x}}{2} = \frac{e^x - e^{-x} + e^x + e^{-x}}{2} = \frac{2e^x}{2} = e^x

  10. Substitute the simplified expression back into the function.

    y=ln⁡(ex)y = \ln(e^x) y=xy = x

  11. Differentiate yy with respect to xx.

    dydx=ddx(x)\frac{dy}{dx} = \frac{d}{dx}(x) dydx=1\boxed{\frac{dy}{dx} = 1}

    : Find the derivative of y=eexy = e^{e^x}.

    We use the chain rule. The derivative of eue^u is eududxe^u \frac{du}{dx}.

  12. Identify the outer and inner functions.

    Let u=exu = e^x.

  13. Differentiate uu with respect to xx.

    dudx=ddx(ex)=ex\frac{du}{dx} = \frac{d}{dx}(e^x) = e^x

  14. Apply the chain rule.

    dydx=eu⋅dudx\frac{dy}{dx} = e^u \cdot \frac{du}{dx} dydx=eex⋅ex\frac{dy}{dx} = e^{e^x} \cdot e^x dydx=ex+ex\boxed{\frac{dy}{dx} = e^{x+e^x}}

    : Find the second derivative of y=xx2+1y = \frac{x}{x^2 + 1}.

  15. Find the first derivative dydx\frac{dy}{dx} using 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=xu = x and v=x2+1v = x^2 + 1.
    Then u′=1u' = 1 and v′=2xv' = 2x.

    dydx=(1)(x2+1)−(x)(2x)(x2+1)2\frac{dy}{dx} = \frac{(1)(x^2 + 1) - (x)(2x)}{(x^2 + 1)^2} dydx=x2+1−2x2(x2+1)2\frac{dy}{dx} = \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2} dydx=1−x2(x2+1)2\frac{dy}{dx} = \frac{1 - x^2}{(x^2 + 1)^2}

  16. Find the second derivative d2ydx2\frac{d^2y}{dx^2} by differentiating 1−x2(x2+1)2\frac{1 - x^2}{(x^2 + 1)^2} using the quotient rule again.

    Let p=1−x2p = 1 - x^2 and q=(x2+1)2q = (x^2 + 1)^2.
    Then p′=−2xp' = -2x.
    To find q′q', use the chain rule: $q' = 2(x^2 + 1) \cdot \frac{d}{dx}(x^2 + 1)

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