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Calculus · real student question

Find dy/dx for x sec y = 4y - y tan 2x.

Question

Find dydx\dfrac{dy}{dx} for

xsec⁡y=4y−ytan⁡2xx\sec y = 4y - y\tan 2x

Step-by-step solution

  1. Treat yy as a function of xx throughout. The equation cannot be solved for yy in closed form, so we differentiate both sides with respect to xx and let every yy carry a hidden dydx\dfrac{dy}{dx} from the chain rule. Both sides contain products of an xx-factor with a yy-factor, so the product rule appears twice.

  2. Differentiate the left side. With u=xu=x and v=sec⁡yv=\sec y, and using ddxsec⁡y=sec⁡ytan⁡y dydx\dfrac{d}{dx}\sec y=\sec y\tan y\,\dfrac{dy}{dx}: ddx(xsec⁡y)=sec⁡y+xsec⁡ytan⁡y dydx.\frac{d}{dx}\left(x\sec y\right)=\sec y + x\sec y\tan y\,\frac{dy}{dx}.

  3. Differentiate the right side, keeping the inner factor of 22. The term 4y4y gives 4dydx4\dfrac{dy}{dx}. For ytan⁡2xy\tan 2x the product rule plus the chain rule (note ddxtan⁡2x=2sec⁡22x\dfrac{d}{dx}\tan 2x = 2\sec^{2}2x, not sec⁡22x\sec^{2}2x) gives ddx(ytan⁡2x)=dydxtan⁡2x+2ysec⁡22x,\frac{d}{dx}\left(y\tan 2x\right)=\frac{dy}{dx}\tan 2x + 2y\sec^{2}2x, so the whole right side differentiates to 4dydx−dydxtan⁡2x−2ysec⁡22x.4\frac{dy}{dx}-\frac{dy}{dx}\tan 2x-2y\sec^{2}2x. Dropping that factor 22 is the single most common error in this problem.

  4. Collect the dydx\dfrac{dy}{dx} terms on one side. Equating the two derivatives, sec⁡y+xsec⁡ytan⁡y dydx=4dydx−dydxtan⁡2x−2ysec⁡22x,\sec y + x\sec y\tan y\,\frac{dy}{dx}=4\frac{dy}{dx}-\frac{dy}{dx}\tan 2x-2y\sec^{2}2x, and moving all dydx\dfrac{dy}{dx} terms right and everything else left: sec⁡y+2ysec⁡22x=dydx(4−tan⁡2x−xsec⁡ytan⁡y).\sec y + 2y\sec^{2}2x=\frac{dy}{dx}\left(4-\tan 2x-x\sec y\tan y\right).

  5. Divide to isolate the derivative. dydx=sec⁡y+2ysec⁡22x4−tan⁡2x−xsec⁡ytan⁡y.\frac{dy}{dx}=\frac{\sec y + 2y\sec^{2}2x}{4-\tan 2x-x\sec y\tan y}. The result legitimately depends on both xx and yy, which is normal for an implicit curve: to get a number you need a point that actually satisfies the original equation.

  6. Check against the implicit function theorem. Writing F(x,y)=xsec⁡y−4y+ytan⁡2xF(x,y)=x\sec y-4y+y\tan 2x, we have Fx=sec⁡y+2ysec⁡22xF_{x}=\sec y+2y\sec^{2}2x and Fy=xsec⁡ytan⁡y−4+tan⁡2xF_{y}=x\sec y\tan y-4+\tan 2x, and dydx=−FxFy\dfrac{dy}{dx}=-\dfrac{F_{x}}{F_{y}} reproduces the same expression. A numerical test confirms it: at x=0.3x=0.3 the curve passes through y=0.09084883y=0.09084883, and a centred difference on the implicitly solved y(x)y(x) gives 0.386471610.38647161, while the formula gives 0.386471610.38647161.

Answer

dydx=sec⁡y+2ysec⁡22x4−tan⁡2x−xsec⁡ytan⁡y\frac{dy}{dx}=\frac{\sec y+2y\sec^{2}2x}{4-\tan 2x-x\sec y\tan y}

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