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The function is bounded on t from $0$ to $2\pi$ And $\vert x \vert \le M$

See that $\vert f(t,x)-f(t,x')\vert \le L\vert x-x'\vert$

I have taken the derivative with respect to x, but don't know what to do from here, does it suffice to say that the Lipschitz constant will be an arbitrary M^3?

Any help regarding multivariate Lipschitz functions would be appreciated

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    Do you understand why it is useful to consider the partial derivative with respect to $x$? If not: take a look at the Mean Value Theorem. It tells you what you have to do, to arrive at a Lipschitz constant. – 2017-02-27

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we have $$f(t,x_2)-f(t,x_1)=x_2(\sin(t)-x_2^2)-x_1(\sin(t)-x_1^2)=\sin(t)(x_2-x_1)-(x_2^3-x_1^3)=\sin(t)(x_2-x_1)-(x_2-x_1)(x_2^2+x_1x_2+x_1^2)$$ does this help you? thus we have $$|f(t,x_2)-f(t,x_1)|\le |x_2-x_1|\left(\left|\sin(t)\right|+|x_2^2+x_1x_2+x_1^2|\right)$$

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    So then this $\vert f(t,x1)-f(t,x2)\vert \le L \vert x1 -x2 \vert$ right? And the $x1$ and $x2$ can be divided out to give the polynomial and the sin function. The sine function is bounded by 1, would you then evaluate $x1$ as it approaches $x2$? – 2017-02-27
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    for $x$ you have $$|x|\le M$$ – 2017-02-27
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    the Lipschitz constant must be a number – 2017-02-27
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    Taking the partial derivative with respect to x then would yield $-4x^2$ as $x1$ approaches $x2$, thus would we have $\vert 4*M\vert \le L$ for the constant? – 2017-02-27
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    i would write $$(1+3M)$$ with $$M>0$$ – 2017-02-27