# Tagged Questions

Questions on the evaluation of derivatives or problems involving derivatives (for example, use of the mean value theorem).

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### Prove that $e^{2x+1}+e^{x}+1/4$ is always greater than 0

Prove that $e^{2x+1}+e^{x}+1/4$ is always greater than 0 I was thinking of supposing that the function is equal to or less than 0, and showing that the supposed equation would not work out. How would ...
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### Why does $\mathrm{d}l=v\,\mathrm{d}t$ imply that $\frac{\mathrm{d}}{\mathrm{d}t}=v\frac{\mathrm{d}}{\mathrm{d}l}$?

If we have a lengthening pendulum and the length $l$ of the pendulum at time $t$ is $$l=l_0+vt$$ where $l_0$ is the initial length of the pendulum and $v$ is the velocity for which the pendulum's ...
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### Can chain rule be used in first step

I was wondering if it were possible to use the chain rule in the first step to differentiate the f.f.g function: $$f(x) = (1 + \sin x)^{\cot x}$$ I know the obvious first step is to use the power ...
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### Can the nabla symbol be used for a covariant derivative?

Can $$\nabla_{\nu} A^{ \mu\nu}$$ represent a covariant derivative with respect to $\nu$? If not what can it be? I'm reading a textbook on General Relativity, and such operations appear without any ...
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### Derivative of convolution type integral equation with respect to time $x(t) = \int\limits_{0}^t \gamma^{t-s} u(s)ds$

Consider the equation: $$x(t) = \int\limits_{0}^t \gamma^{t-s} u(s)ds$$ where $x, u$ are functions and $\gamma \in \mathbb{R}_{>0}$ I am trying to obtain the derivative of this equation with ...
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### Total derivative using $(x,y)\mapsto f(x,y)$ notation

Related to What is the difference between $\frac{\mathrm{d}}{\mathrm{d}x}$ and $\frac{\partial}{\partial x}$? Is it possible to make sense of $(x,y)\mapsto f(x,y)$ when taking the total derivative ...
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### Prove that $\sin x \cdot \sin (2x) \cdot \sin(3x) < \tfrac{9}{16}$ for all $x$
Prove that $$\sin (x) \cdot \sin (2x) \cdot \sin(3x) < \dfrac{9}{16} \quad \forall \ x \in \mathbb{R}$$ I thought about using derivatives, but it would be too lengthy. Any help will be ...