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0

The rule is $$\frac d{dx}u(x)^{v(x)}=v(x)u(x)^{v(x)-1}\frac{du}{dx}+u(x)^{v(x)}(\ln u(x))\frac{dv}{dx}.$$ You can derive this by the method of logarithmic differentiation. It is a special case of the two-variable chain rule: $$\frac{dw}{dx}=\frac{\partial w}{\partial u}\frac{du}{dx}+\frac{\partial w}{\partial v}\frac{dv}{dx}.$$ Notice how it simplifies to ...

5

I will assume that this is not homework and go ahead to give a full solution. As others have said, your rule only works for constant $a$. Following the hint in danielson's answer and setting $y=f(x)$, you have $$\log y=\sin x\log\left(1+x^2\right)$$ and differentiating both sides with respect to $x$ and using the chain and product rules gives ...

3

Start with $$f(x)=e^{\ln(x^2+1)\sin(x)}$$ The chain rule gives $$f'(x)=e^{\ln(x^2+1)\sin(x)}\times (\ln(x^2+1)\sin(x))'=(x^2+1)^{\sin(x)}(\ln(x^2+1)\sin(x))'$$ Take it from here.

2

The rule that you have given is for $\textit{constant}$ $a$, while the base of your function is a $\textit{function}$. So the rule does not apply here. To fix this, let $f(x) = (x^2 + 1)^{\sin(x)}$ and consider $\ln(f(x))$. Differentiate the new function, while also utilizing a nice property of exponents inside logarithmic functions, and try to deduce the ...

7

The rule that you quote is only valid if a is a constant.

0

For $x\in\mathbb{R}$ we have $$e^{ix}= \cos x+ i \sin x,$$ $$\cos ix= \frac{e^{-x}+ e^{x}}{2}=\cosh x,$$ and $$\sin i x= \frac{e^{-x}- e^{-x}}{2i}=i\sinh x$$ Thus, $$\cos ix+ i \sin ix=\cosh x-\sinh = \frac{e^{x}+ e^{-x}}{2} - \frac{e^{x}- e^{-x}}{2}=e^x$$ so, $$\sin ix = i({\cos ix- e^x})= i({\cosh x- e^x})=i\sinh x,$$ i.e. $\sin ix$ is an imaginary ...

0

First, note that $$\mbox{for }y\in\mathbb{R}\ \mbox{we have}\ \sin{y}=\mathrm{Im}\{e^{iy}\}.$$ Stating $\mathrm{Im}\left\{\mathrm{exp}(-x)\right\}=0$ means you assume $x\in\mathbb{R}$. However, then $ix$ is complex, such that you cannot apply the above rule.

3

This, of course, uses three interconnected formulas: $e^{ix}= cos(x)+ i sin(x)$, $cos(x)= \frac{e^{ix}+ e^{-ix}}{2}$, and $sin(x)= \frac{e^{ix}- e^{-ix}}{2}$ Your error is that you are assuming that the imaginary part of $e^{ix}$ is "i sin(x)". That is true only if itself is real. If x is not real the $i sin(x)$ is not imaginary because sin(x) is not ...

0

Usually Laplace transform is defined as $$\mathcal{L} f= \int_{0}^{+\infty} f(t) e^{-st}dt$$ The transform $${\mathcal{B}} f= \int_{-\infty}^{+\infty} f(t) e^{-st}dt$$ is a bilateral Laplace transform. For a Laplace transform $\mathcal{L} e^{at}f = F(s-a)$ where $F=\mathcal{L} f$, i.e. a frequency shifting. Similarly $\mathcal{L} e^{-at}f = F(s+a)$ ...

0

$u^{\frac12 + \frac14 + \frac 18 + \cdots} \rightarrow u$ for all $u$ and $2^n \rightarrow \infty$ so the bound is infinite if $u > 0$ and $1+0$ if $u = 0$.

2

Apply the squeeze theorem: $$\lim_{x\to\infty}(e^x)^{\frac1x}\leq \lim_{x\to\infty}(e^x+1)^{\frac1x}\leq \lim_{x\to\infty}(e^x\cdot e)^{\frac1x}$$

2

Write: $$(e^x+1)^{1/x}=(e^x)^{1/x} (1+e^{-x})^{1/x}=e\times (1+e^{-x})^{1/x}$$ which goes to $e$ as $x \to \infty$

3

So we have $x^2 = 2^x$. Taking the square root of both sides and assume the solution is negative gives $x=-\sqrt{2}^x$. We can then establish a recursive sequence, $x_n = -\sqrt2^{x_n-1}$. Assuming that this converges gives us the answer, $x=-\sqrt2 ^{-\sqrt2 ^ {-\sqrt2 ^\cdots}}$. After five iterations, we get $$x\approx-0.76961847524.$$ Substituting the ...

7

Suppose , $gcd(a,b)=1$ , $x=\frac{a}{b}$ and $x^2=2^x$ We have $$a^2=b^2\times 2^{a/b}$$ implying $$a^{2b}=b^{2b}\times 2^a$$ This is impossible, if $gcd(a,b)=1$ and $b>1$. The negative solution is obviously not an integer. If $x$ is irrational algebraic, then $2^x$ is transcendental, but $x^2$ is not. So, $x$ , the negative solution, must be a ...

0

$$\lim_{\alpha\to 0}\frac{e^{-\sqrt{z(z+r))}}}{1+\alpha\sqrt{z(z+r)}+(1-\alpha \sqrt{z(z+r)})e^{-2\sqrt{z(z+r))}}}=$$ $$\lim_{\alpha\to 0}\frac{1}{e^{\sqrt{z(z+r))}}\left(1+\alpha\sqrt{z(z+r)}+\frac{1-\alpha\sqrt{z(z+r)}}{e^{2\sqrt{z(z+r)}}}\right)}=$$ $$\frac{1}{\lim_{\alpha\to ... 3 Let f(x)=y=\dfrac{e^x-e^{-x}}2=\dfrac{e^{2x}-1}{2e^x} \implies e^{2x}-2e^x y-1=0\implies e^x=\cdots\implies x=\ln(\cdots) \implies f^{-1}(y)=x=\cdots 0 HINTS:$$\mathscr{L}\left(\sin (wt)u(t)\right)(p)=\frac{w^2}{p^2+w^2}\mathscr{L}(f(t-0.5)u(t-0.5))(p)=e^{-0.5p}\mathscr{L}(f(t)u(t))(p)$$1 A less high-powered approach. I am assuming that since this is a model of daily sales that t is the number of days, and it doesn't make sense to push the value of t any further than the nearest integer. So we just want to find the day that sales pass 4500. You went a step too far already. you want to put it in this form:$$200(t -15)e^{-0.01t} + 3000 = ...

1

When you see the variable you are trying to solve for in both the coefficient and exponent, the Lambert W function is always an appropriate path to go. First, lets isolate the first term. $200te^{-0.01t}=1500+3000e$ Now transform the first term so that the coefficient is the same as the exponent. $te^{-0.01t}=7.5+15e$ $-0.01te^{-0.01t}=-0.075-0.15e$ ...

1

This question could have been phrased better, but nevertheless I'll give a quick explanation. $$(3y^5)^6=3^6(y^5)^6=729y^{30}.$$ This follows from elementary rules of exponentiation. You should check these in your math text.

1

If you know that the function $\mathbb{R} \to \mathbb{R}, x\mapsto x^{\frac{p}{q}}$ is continuous for $p,q \in \mathbb{Z}$, $q\ne 0$ then you can rewrite $f$ as $$f\left( \frac{p}{q} \right)= \left( \lim_{n\to \infty} \left(1+\frac{1}{n}\right)^n \right)^{\frac{p}{q}} = \lim_{n\to \infty} \left(1+\frac{1}{n}\right)^{\frac{np}{q}} .$$ Now you can also look ...

1

One way to do this is to define $$L(x) = \int_1^x \frac 1t \, dt, \quad x > 0$$ and show that $f$ is differentiable and monotone on $(0,\infty)$ with range $(-\infty,\infty)$. Define $E(x) = L^{-1}(x)$. It is an exercises in calculus to deduce the usual rules of logarithms and exponents. Show that $g(x) = E(x)$ by showing that $\lim_{n \to\infty} ... 5 I don't know what you mean by "infinite number of integrals", but we can do something like the following: Let$T:C(\mathbb{R})\rightarrow C(\mathbb{R})$be an operator on the space of continuous functions on$\mathbb{R}$such that for$f\in C(\mathbb{R})$, the function$Tf\in C(\mathbb{R})$is defined by $$(Tf)(x) = 1+\int\limits_{0}^{x}{f(t)\text{ d}t}. ... 4 Here's a beautiful relation between \pi and e,$$\sqrt{\frac{\pi\,e}{2}}=1+\frac{1}{1\cdot3}+\frac{1}{1\cdot3\cdot5}+\frac{1}{1\cdot3\cdot5\cdot7}+\dots+\cfrac1{1+\cfrac{1}{1+\cfrac{2}{1+\cfrac{3}{1+\ddots}}}}$$It shouldn't be hard to guess who found this. As Kevin Brown of Mathpages remarked, "Is there any other mathematician whose work is instantly ... 4 I honestly just like the fact that e + \pi might be rational. This is the most embarrassing unsolved problem in mathematics in my opinion. It's clearly transcendental and we have no idea how to prove that it's even irrational. They're so unrelated additively that we can't prove anything about how unrelated they are. Uh-huh. e\pi might also be rational. ... 7$$e^{\pi\sqrt{163}} =262537412640768743.99999999999925...$$1 You may consider expressing Euler's identity as e^{i\pi}+1 = 0 instead of the way you have it, because 0 shows up. 14$$\int_{-\infty}^\infty\frac{\cos x}{x^2+1}\operatorname d\!x=\frac\pi e$$EDIT: Also:$$\int_{-\infty}^\infty e^{-x^2}\operatorname d\!x=\sqrt\pi$$6 Some additional possibilities$$n!\sim\sqrt{2\pi n}\left(\frac {n}{e}\right)^n$$The normal distribution is given by$$\phi(x) = \frac{1}{2 \pi}e^{(-1/2)x^2}\int_{-\infty}^\infty\phi(x)dx=1$$A personal favorite involving the Euler–Mascheroni constant.$$\int_0^\infty e^{-x}\ln^2x \,dx=\gamma^2+\frac{\pi^2}{6}$$Also ... 13 I'm a fan of$$e^\pi-\pi=20{}$$(Well... almost...) 1 First you set the entire equation equal to y:$$ (\frac {x}{(x+1)})^x = y $$We can then insert both sides into ln(x):$$ ln((\frac x{(x+1)})^x) = ln(y) $$Then pull out the x power:$$ xln(\frac x{(x+1)}) = ln(y) $$Split the natural log:$$ xln(x) - xln(x+1) = ln(y) $$Shift the left term into the bottom of a fraction and add it to the other ... 2 Another way to get it. Consider$$A=\left(\frac{x}{x+1}\right)^x$$and take logarithms$$\log(A)=x \log\left(\frac{x}{x+1}\right)=x \log\left(\frac{x+1-1}{x+1}\right)=x \log\left(1-\frac{1}{x+1}\right)$$Now, remember that, for small y, \log(1-y)\approx -y. Replace y by \frac{1}{x+1} which makes$$\log(A)\approx -\frac{x}{x+1}=-\frac{1}{1+\frac ... 2 There are a variety of approaches to evaluate the limit of interest. I thought that it would be instructive to show a way forward that uses only the squeeze theorem and standard inequalities. To that end we proceed. I showed in THIS ANSWER that the logarithm function satisfies the inequalities for$z>0$$$\frac{z-1}{z}\le \log z \le z-1 \tag 1$$ ... 1$\lim_{x\to\infty} (\frac{x}{(x+1)})^x\\ =\lim_{x\to\infty}(1-\frac{1}{(x+1)})^{(x+1)-1}\\ =\lim_{x\to\infty}(1-\frac{1}{(x+1)})^{(x+1)}\lim_{x\to\infty}(1-\frac{1}{(x+1)})^{-1}\\ =e^{-1}\cdot 1\\ =e^{-1}Notice you should transform your expression to well-known form. 8 Let's start with finding the limit of the reciprocal of the original expression: \begin{align} & \left(1 + \frac{1}{x}\right)^x \\ \end{align} whose limit is the well-knowne. Now you can conclude easily. For those who needed more details, let me put it more explictly: $$\left(\frac{x}{x + 1}\right)^x = \left(\frac{1}{1 + \frac{1}{x}}\right)^x = ... 0 8n^2=64n log_2(n)\iff\frac n8= log_2 (n)\iff 2^{\frac n8}=n So we have (\sqrt[8] 2)^n=n. This is a trascendental equation which can be solved by approximations or by computer (which is approximated too). I rather try the approximation by hand in this particular case. \alpha=\sqrt[8] 2=1.090507733 so one has \alpha^n=n to solve. It is clear that ... 1 You have a polynomial equation in n and log_{2}(n). For all n\neq 1, the both are algebraic independent. That means, n is transcendental. Use the Lambert W function. Maple gives n=-\frac{8LambertW(-\frac{1}{8}ln(2))}{ln(2)},-\frac{8LambertW(-1,-\frac{1}{8}ln(2))}{ln(2)}. Type in Wolfram Alpha: Solve(8*n^2=64*n*log2(n),n). 1$$I=\frac{\text{d}}{\text{d}m} \int_{x=-\infty}^m\int_{y=n}^{+\infty}\exp\left(-\left[\left(\frac{x-a}{b}\right)^2 - \frac{(x-a)(y-a)}{a}+\left(\frac{y-a}{c}\right)^2\right]\right)\space\text{d}y\text{d}x$$Mathematica gives a closet form: ... 1 Ian say that \frac{d}{dx} \int_a^x f(u) du = f(x) whatever a fixed. so \frac{d}{dm} \int_{x=-\infty}^m\int_{y=n}^{+\infty}\exp\left(-\left[\left(\frac{x-a}{b}\right)^2 - \frac{(x-a)(y-a)}{a}+\left(\frac{y-a}{c}\right)^2\right]\right)\space dy dx=\int_n^{\infty}\exp\left(-\left[\left(\frac{m-a}{b}\right)^2 - ... 3 Naturally, you can take derivative \partial^2_{a,b} M_{X,Y}(0,0) to get the answer. Alternatively, you can interpret the joint distribution of X and Y as a mixture. Namely, a random variable \Lambda takes value 1 with probability 4/5 and value 2 with probability 1/5. Given \Lambda = \lambda, X and Y are independent and exponentially ... 1 You're supposed to take the definite integral over the entire time domain$$ \int_0^\infty \left(e^{-t/\tau_1}-e^{-t/\tau_2}\right)\, dt = \left[\tau_1e^{-t/\tau_1} - \tau_2e^{-t/\tau_2}\right]_0^\infty = \tau_1 - \tau_2 1 The growth of x^n is called linear (n=1), quadratic (n=2), cubic (n=3), quartic (n=4), quintic (n=5)... You can find here a few names after that, but I don't expect anyone to say "octavic/octic" with a straight face. In general if you don't know n it's just called "polynomial growth". 0 For exponent 2, you say quadratic, for 3 and following, cubic, quartic, quintic... For general exponent, power function or power law. [In French you can use potentielle, but this can cause polysemic ambiguities.] The adverb quadratically can be freely used; I would abstain from "cubicly", "quarticly"... Also avoid the confusion with quadric, which ... 0 Suppose t_0 = kT + \alpha for some \alpha \in [0, T), k \in \mathbb{Z} \begin{align} &\int_0^T \tilde{K} e^\tau I(\tau + t_0) \mathrm{d}\tau \overset{(a)}= \int_0^{T} \tilde{K}e^\tau I(\tau + \alpha) \mathrm{d}\tau \\ = ~& \int_0^{T-\alpha} \tilde{K} e^\tau I(\tau + \alpha) \mathrm{d}\tau + \int_{T-\alpha}^T \tilde{K} e^\tau I(\tau + \alpha) ... 0\frac{d}{dx}\left[e^\sqrt x + \ln\sqrt x\right]=\frac{d}{dx}\left[e^\sqrt x\right] + \frac{d}{dx}\left[\ln\sqrt x\right]=e^\sqrt{x}\frac{d}{dx}\left[\sqrt{x}\right] + \frac{1}{\sqrt{x}}\frac{d}{dx}\left[\sqrt x\right]=\frac{d}{dx}\left[\sqrt x\right]\left(e^\sqrt{x}+ \frac{1}{\sqrt{x}}\right)=\frac1{2\sqrt{x}}\left(e^\sqrt{x}+ ... 0 Since $$D_{x}(\sqrt{x}) = \frac{1}{2 \, \sqrt{x}}$$ then \begin{align} D_{x} y(x) &= D_{x} \left[ e^{\sqrt{x}} + \ln(\sqrt{x}) \right] \\ &= e^{\sqrt{x}} \, D_{x}(\sqrt{x}) + \frac{1}{\sqrt{x}} \, D_{x}(\sqrt{x}) \\ &= \frac{1}{2 \, \sqrt{x}} \, \left[ e^{\sqrt{x}} + \frac{1}{\sqrt{x}} \right] \end{align} 0 No, but you can use the chain rule: $$(g(f(x))' = g'(f(x))f'(x).$$ So $$(e^{\sqrt{x}})' = e^{\sqrt{x}}\frac{1}{2\sqrt{x}}$$ and for the other one, you can either use the chain rule again, or you can use that\ln(\sqrt{x}) = \frac{1}{2}\ln(x)$. 0 Not quite, you want the chain rule. The first one would have the form$e^u \mapsto u'e^u.$In this case, that's$\frac{1}{2 \sqrt{x}}e^\sqrt{x},$by just taking the derivative of$\sqrt{x}$. Can you do the other one? 0 Hint: Take$x=\dfrac1k$and use the definition of$e$: $$\lim_{x\to0}\frac{e^x-1}x=\lim_{k\to\infty}k(e^{1/k}-1)=\lim_{k\to\infty}k\left(\lim_{nk\to\infty}\left(1+\frac1{nk}\right)^{nk/k}-1\right).$$ By the Binomial theorem, the inner limit expands as $$1+\frac n{nk}+\frac{n(n-1)}{2n^2k^2}+\frac{n(n-1)(n-2)}{3!n^3k^3}\cdots,$$ which is bounded above by ... 4 Set$\ e^x-1=y, $so$\ x=\ln(1+y)$and$\ y\to0$for$\ x\to0$Now you have: $$\ \lim_{x\to0}\frac{e^x-1}{x}=\lim_{y\to0}\frac{y}{\ln(1+y)}=\lim_{y\to0}\frac{1}{\frac{\ln(1+y)}{y}}=\lim_{y\to0}\frac{1}{\ln(1+y)^{\frac{1}{y}}}$$ Putting$\ \frac{1}{y}=t,$for$\ y\to0, t\to\infty$, so you have:$\$\ ...

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