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I always had the same weird feeling when I solve elementary differential equations or differentiate implicit functions. But the following will resolve your concerns thoroughly. If you have plotted the graph of $\cos x + \cos y = 1/2$, you could see that the derivative at $\sin y = 0$ is actually undefined (vertical). Therefore you can only differentiate ...

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You have already obtained a relation between $x,y,y'$. Now, you need to "solve" with respect to $y'$, getting $$y' = \dfrac{-6x^2 y-5}{2x^3+6-\pi \cos(\pi y)}.$$

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This is a pretty straightforward application of the chain rule. Given \begin{aligned} 0 &= f(\lambda, w(\lambda), s(\lambda)) = (1-\lambda s) w - 1 \\[2ex] \implies 0 &= \frac{{\rm d\,} f(\lambda, w(\lambda), s(\lambda))}{{\rm d\,} \lambda} = \frac{\partial f}{\partial \lambda} + \frac{\partial f}{\partial w}\frac{\partial w}{\partial \lambda} + \... 1 If we write it as \frac{dy}{dx}-y=-x \, , $$then we are dealing with a linear first-order differential equation of the form$$ \frac{dy}{dx}+p(x)y=q(x) \, . $$To solve an equation of this form, you have to multiply both sides by e^{P(x)}, where P is an antiderivative of p. Here p(x)=-1, and so we can take P(x)=-x. The equation becomes$$ e^{-x}...

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One can find a rigorous definition in [Wikipedia][1] , however I will attempt to provide a practical way of thinking of implicit function's. An explicit single variable function can be thought of as a 'mapping' from one set to another set of the form $x \to y$, this can be graphed on a cartesian grid by highlighting all the pair $(x,y)$ included in this ...

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We could use $$y = \sqrt{r^2-x^2}$$ $$y = -\sqrt{r^2-x^2}$$ For a circle of radius $r$ centered at $O$, with the equations representing the top and bottom semi-circles respectively.

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For ease of typing, define \eqalign{ G &= \Gamma = (I+wC) \quad\qquad F=(BGB^T)^{-1} \quad\qquad p=(B^Th-a) } and note that $F$ and $G$ are symmetric since $C^T=C.$ Next, eliminate the variable $s$ in favor of $w$ \eqalign{ s &= p^TCp \\ w &= (1-\lambda s)^{-1} \;\doteq\; (1-\lambda p^TCp)^{-1} \\ \tfrac{w-1}{w\lambda} &= p^TCp \\ } ...

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Careful, I edited what I thought was a typo, but it turned out to be the source of your error, so I reversed the edit. $F(x,y) = 10$ when $(x,y) = (1,\ln(11))$, not when $(x,y) = (1,\log(11))$, where I (and your calculator) are using $\log(11)$ to mean the base-$10$ logarithm, not the base-$e$ logarithm. Both your derivatives are correct, and when you plug ...

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