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 Nov 10 comment Using L'Hopital to solve $\lim_{x\to +\infty}\frac{\frac{-1}{x^2}}{\sin^2\left(\frac{2}{x}\right)}$ Why did you pick $\frac{2}{x}$ in particular? I mean, I can see why your solution works, but I'm having a hard time understanding your reasoning behind picking $\frac{2}{x}$ and not something else. Is it because of the sine function? Nov 10 comment Using L'Hopital to solve $\lim_{x\to +\infty}\frac{\frac{-1}{x^2}}{\sin^2\left(\frac{2}{x}\right)}$ I can see and understand why this works. However, it's not clear to me what triggered your idea of dividing $\frac{-1}{x^2}$ by $\frac{-1}{4}$. What made you see that? Nov 9 comment Solving $\lim_{x\to-\infty}x^2\cdot e^x$ with L'Hopital Do we have to change $x\to -\infty$ to $x \to \infty$? Can't we just keep $e^{-x}$? Because when you evaluate, it will become $e^{--\infty}$ anyway. Nov 9 comment Applying L'Hopital on $\lim_{x\to1^+}\left( \frac{1}{\ln(x)} - \frac{1}{x - 1} \right)$ So $0^+ - 0^+$ is indeed $0$? And $0^+ \cdot 0^+$ is also $0$? I always thought that $0^+$ meant a number greater than $0$. Nov 9 comment Applying L'Hopital on $\lim_{x\to1^+}\left( \frac{1}{\ln(x)} - \frac{1}{x - 1} \right)$ Well, it's good to know I can do this, but I suspect I'm not supposed to know / use this yet. Do you know if $0^+ - 0^+ = 0$? Or if $0^+ \cdot 0^+ = 0$? Nov 9 comment Getting the rate of change of a rectangle's base if its top-right corner is defined as $y = 2^x$ You... you should've said that instead! Nov 9 comment Getting the rate of change of a rectangle's base if its top-right corner is defined as $y = 2^x$ Hmm... So $$y = e^{x \cdot \ln(2)}$$ If I want to know the rate of change of $x$ I would need... the derivative, I guess. Which is $$\frac{dy}{dt} = e^{x \cdot \ln(2)}\cdot \ln(2) \cdot \frac{dx}{dt}$$ Since $\frac{dy}{dt} = 1$ I can have that $$1 = e^{x\cdot \ln(2)}\cdot \ln(2) \cdot \frac{dx}{dt} \implies \frac{dx}{dt} = \frac{\frac{1}{\ln(2)}}{e^{x\cdot \ln(2)}}$$ Then replace $x = 2$ and we get that $$\frac{dx}{dt} = 0.36$$This seems to lead me to the correct answer according to the book. Thank you. Nov 9 comment Related rates: using angle to find rate of change of opposite side Ah, I see my problem. Thanks. However the answer is $$\frac{13\pi}{60}$$ Nov 9 comment Related rates: derivative of the function $A = \frac{x\cdot y}{2}$ Indeed, that's exactly it. I had interpreted $$\frac{x\cdot y}{2} = \frac{x}{2}\cdot\frac{y}{2}$$ when it should've been $$\frac{x\cdot y}{2} = \frac{x}{2} \cdot y$$ Thank you. Nov 9 comment Related rates: derivative of the function $A = \frac{x\cdot y}{2}$ Hm, I just confirmed that I have an extra $\frac{1}{2}$ there. But I don't really get it because the derivative of $\frac{x}{2}$ is definitely $\frac{1}{2}\cdot \frac{dx}{dt}$ and then multiplied by the second part of the product, which is $\frac{y}{2}$. Nov 9 comment Related rates: derivative of the function $A = \frac{x\cdot y}{2}$ Hmm... But if I have $\frac{x}{2}$, whose derivative is $\frac{1}{2} \cdot \frac{dx}{dt}$, by the product rule I have to multiply this by the $\frac{y}{2}$ (it's the other side of the product) which results in $$\frac{1}{2}\cdot\frac{dx}{dt}\cdot\frac{y}{2}$$. Nov 9 comment Related rates: derivative of the function $A = \frac{x\cdot y}{2}$ What's the problem with my attempt at the product rule though? Nov 9 comment Related rates: derivative of the function $A = \frac{x\cdot y}{2}$ I don't quite see why are you multiplying $\frac{dx}{dt}$ by $\frac{dA}{dx}$. Why $\frac{dA}{dx}$? Nov 9 comment Help understand related rates problem: calculating the derivative of the distance function So let's see if I get this straight: $z$, $x$, $y$ are all functions of $t$ (and they are actually $z(t)$, $x(t)$ and $y(t)$), so $$[z(t)]^2$$ would have as derivative $$2\cdot z(t)$$ and by the chain rule, I have to multiply this by the derivative of $z(t)$ (which is $\frac{dz}{dt}$) and therefore the full derivative of $[z(t)]^2$ is $$2\cdot z(t) \cdot \frac{dz}{dt}$$? Nov 8 comment Find the equations to the normal lines of a curve where $x = 4$. Oh! Right! The normal line. I got it now. Thank you! Nov 8 comment Find the equations to the normal lines of a curve where $x = 4$. I see. $\frac{5}{10}$ would be the slope for a tangent line that passes by $(4,3)$. The tangent line equation would be $$y-3=\frac{5}{10}(x-4)$$ Which is $$y=\frac{x}{2}+1$$ It concerns me, because apparently the answer should be $$y=-2x+11$$ According to my book. Nov 8 comment Find the equations to the normal lines of a curve where $x = 4$. I have calculated the implicit derivative to be $$\frac{2x-y}{2y+x}$$ So the slopes of the tangent lines at $x = 4$ should be obtained by evaluating $x=4$ in this derivative, rather than the original function $y^2+4y-16=5$, no? If I plug $x=4$ in this derivative, I would get that $y=0$. Nov 8 comment Find the equations to the normal lines of a curve where $x = 4$. @G-man I mean getting the derivative of $y^2+4y-16=5$. I assumed that you could calculate derivatives of functions only. Nov 8 comment Find the equations to the normal lines of a curve where $x = 4$. @G-man ah, well, I didn't know that. So this curve is not a function? Does that mean I can't derive it? (I was guessing deriving it would be part of the solution or something like that) Nov 3 comment Calculate the derivative of the product of three functions $e^x\cdot \ln(x) \cdot \cot x$ @Kaster that looks pretty cool - how did you reach that conclusion?