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Comments on Consider the second of these integrals (What's the meaning of second right here?)

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Consider the second of these integrals (What's the meaning of second right here?)

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$$\frac{dJ}{d\alpha}=\int_{x_1}^{x_2}(\frac{\partial f}{\partial y}\frac{\partial y}{\partial \alpha}+\frac{\partial f}{\partial \dot{x}}\frac{\partial \dot{x}}{\partial \alpha})\mathrm dx$$ Consider the second of these integrals: $$\int_{x_1}^{x_2}\frac{\partial f}{\partial \dot{y}}\frac{\partial \dot{y}}{\partial \alpha}\mathrm dx=\int_{x_1}^{x_2}\frac{\partial f}{\partial \dot{y}}\frac{\partial^2 y}{\partial x \partial \alpha}\mathrm dx$$

What did they mean by "second"? There must be a negative on LHS or RHS. What happened to the line? To me, it seems like he had used chain rule for $\partial y$. But, why there's no negative?

picture of book used double blockquote to decrease size of the picture

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In the post, actually $\dot{y}=\frac{dy}{dx}$. So,

$$\int_{x_1}^{x_2}\frac{\partial f}{\partial \dot{y}}\frac{\partial \dot{y}}{\partial \alpha}\mathrm dx=\int_{x_1}^{x_2}\frac{\partial f}{\partial \dot{y}}\frac{\partial^2 y}{\partial x \partial \alpha}\mathrm dx$$

In the equation, they just wrote $\frac{dy}{dx}$ instead of $\dot{y}$

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Yes, that's what I said in my comment yesterday. (2 comments)
Yes, that's what I said in my comment yesterday.
Peter Taylor‭ wrote about 3 years ago

Yes, that's what I said in my comment yesterday.

deleted user wrote about 3 years ago · edited about 3 years ago

Maybe, I didn't understand you properly. I started reading from above then I got where the error was.. Anyway, Thanks