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I have the following function $$f^{(0)}\left(x\right)=f\left(x\right)=e^{x}$$ And want to approximate it using Taylor at the point ##\frac{1}{\sqrt e} ## I also want to decide (without calculator)whether the error in the approximation is smaller than ##\frac{1}{25} ## The Taylor polynomial is...

for all real numbers except for 0 right?

Hi there. I have the following function: $$f(x)=\ln|\sin(x)|$$ I've caculated the derivative to: $$f'(x)=\frac{\cos(x)}{\sin(x)}$$ And the domain of f(x) to: $$(2\pi n, \pi+2\pi n ) \cup (-\pi + 2\pi n, 2\pi n)$$ And the domain of f'(x) to: $$(\pi n, \pi+\pi n )$$ I want to determine for...

I understand :) Thank you! Is it enough "proof" to show the derivative exists by calculating the domain of the derivative? or is there other ways?

the derivative at 0 is undefined right?

But what I have hard to understand is that the function is defined at [0,1] but does the derivative exists at the endpoints?

Hi there. I have the following function: $$f(x)=arcsin(\sqrt x)$$ I've caculated the derivative to: $$f'(x)=\frac{1}{2\sqrt x\sqrt{ (1-x}}$$ And the domain of f(x) to: $$[0, 1]$$ And the domain of f'(x) to: $$(0, 1)$$ I want to determine for which x the derivative exists but I'm not...

I really don't know.. are you reffering to the extremum point at x=-2?

Hi there. I have the following function: $$f(x)=x+\frac{1}{(x+1)}$$ I've caculated the derivative to: $$f'(x)=1-\frac{1}{(1+x)^2}$$ And the domain to: $$(-\infty, -1)\cup(-1, \infty)$$ I've also found two extreme point: $$x=0, x=-2$$ I know that a function is strictly increasing if...

The vector field F which is given by $$\mathbf{F} = \dfrac{(x, y)} {\sqrt {1-x^2-y^2}}$$ And the line integral $$ \int_{C} F \cdot dr $$ C is the path of $$\dfrac{\ (\cos (t), \sin (t))}{ 1+ e^t}$$ , and $$0 ≤ t < \infty $$ How do I calculate this? Anyone got a tip/hint? many thanks

Hahaha.. I finally got it! the mixed partial cancels and if I use the pythagorean trigonometric identity on what's left I get the answer! I can't explain how much you've made my day! <3

then I get: $$\dfrac{\partial^2}{\partial y^2} = \sin^2 \theta \dfrac{\partial^2}{\partial u^2} - 2sin \theta\cos \theta\dfrac{\partial}{\partial u\partial v}+ \cos^2 \theta \dfrac{\partial^2}{\partial v^2}$$ Can I somewhat use this to calculate $$\dfrac{\partial^2}{\partial u^2}$$ ? Thank...

$$\dfrac{\partial^2}{\partial x^2} = \cos^2 \theta \dfrac{\partial^2}{\partial u^2} + 2sin \theta\cos \theta\dfrac{\partial}{\partial u\partial v}+ \sin^2 \theta \dfrac{\partial^2}{\partial v^2}$$ Is that correct? :) Can you give me any tips on how to calculate $$\dfrac{\partial^2}{\partial...

Thank you so much for helping me.. should the first line be: $$\dfrac{\partial}{\partial x} = \dfrac{\partial u}{\partial x}\dfrac{\partial}{\partial u} + \dfrac{\partial v}{\partial x}\dfrac{\partial}{\partial v} = \cos \theta \dfrac{\partial}{\partial u} + \sin \theta\dfrac{\partial}{\partial...

Mentor note: Fixed the LaTeX in the following I have the following statement: \begin{cases} u=x \cos \theta - y\sin \theta \\ v=x\sin \theta + y\cos \theta \end{cases} I wan't to calculate: $$\dfrac{\partial^2}{\partial x^2}$$ My solution for ##\dfrac{\partial^2}{\partial x^2}##...