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16

Do you know the Leibniz formula for derivative $$(fg)^{(n)}=\sum_{k=0}^n{n\choose k}f^{(k)}g^{(n-k)}?$$ Added: Since the third derivative of $x^2$ is $0$ so $$(x^2e^x)^{(10)}=\sum_{k=0}^2{10\choose k}(x^2)^{(k)}(e^x)^{(10-k)}\\={10\choose 0}x^2e^x+{10\choose1}\times 2xe^x+{10\choose 2}\times2e^x=(x^2+20x+90)e^x$$

11

Hint: The derivative of $P(x)e^x$ is $(P'(x)+P(x))e^x$. So, basically, you have to apply 10 times the transformation $P \to P' + P$, which should not be difficult with the polynomial $P(x)=x^2$: $$x^2$$ $$x^2+2x$$ $$(x^2+2x)+(2x+2)=x^2+4x+2$$ $$(x^2+4x+2)+(2x+4)=x^2+6x+6$$ $$(x^2+6x+6)+(2x+6)=x^2+8x+12$$ $$(x^2+8x+12)+(2x+8)=x^2+10x+20$$ The last line is ...

5

Hint $\ \ f = (ax^2\!+bx+c)\,e^x\Rightarrow\, f' = (ax^2\!+(b+2a)x+c+b)\,e^x.\,$ Abbreviating $f\mapsto f'$ as $$(a,b,c)\ \mapsto\ (a, b+2a,c+b)$$ We iterate it $n$ times to get $f^{(n)}.$ Looking for patterns in the first few values we have $$\begin{eqnarray} f^{(1)} = (a,&&b+2a,&&c\, +\, b)\\ f^{(2)} = ... 4 For fun, since good ways have been described, we do not so good. First a small change of notation. Let f(x)=x^2e^x. We find f^{(10)}(a), or more generally f^{(n)}(a). Expand e^x in a power series about x=a. We get$$e^x=\sum_0^\infty \frac{e^a}{n!}(x-a)^n.\tag{1}$$Expand x^2 about x=a. We get$$x^2=a^2+2a(x-a)+(x-a)^2.\tag{2}$$For n\ge ... 4 One trick to compute derivatives of function containing an exponential factor e^{\alpha x} is the formal identity$$e^{-\alpha x} \frac{d}{dx} e^{\alpha x} = \frac{d}{dx} + \alpha$$What this means is if you put any differentiable function f(x) on the RHS on both sides of above formal identity, you get back a valid identity:$$ e^{-\alpha ...

4

The reason you need two integrals is because over the range $0$ to $4$ in $y$, the height of the shell is $2\sqrt y$ because it extends all the way across the $y$ axis. Over the range $4$ to $9$ in $y$, the height is $\sqrt y -(y-6)$. That changes the integrand, so split it into two integrals. A sketch is below. The horizontal lines are what are revolved ...

3

Use the general formula for a differentiable function $\;f(x)\;$, which you can easily prove using the Chain Rule : $$\int \frac{f'(x)}{\sqrt{f(x)}}dx=2\sqrt{f(x)}+C$$ And then, since $\;(x^4-2)'=4x^3\;$ , we get ...

3

The geometric mean of the continuous interval $[a,x]$ (with $a > 0$) is $$GM(a,x) = \frac{1}{e} \left(\frac{x^x}{a^a} \right)^{\frac{1}{x-a}}.$$ Geometric means have applications in several areas, such as finance. The derivation of this formula is a nice exercise in second-semester calculus. It can be done as an application of the integral, perhaps ...

2

Part (a) is just an application of the chain rule: this function is of the form $f(x) = g(h(x))$ where $g(u) = \ln(u)$ and $h(x) = x^2 - 2x + 2$. Then the chain rule informs us: $$f'(x) = g'(h(x)) h'(x) = \frac{1}{h(x)} (2x - 2) = \frac{2x-2}{x^2 - 2x + 2}.$$ For part (b), you need to find the values of $x$ so that $f'(x) = 0$, i.e. for what $x$ is $$... 2 If considered as function of c, the expression is piecewise-linear, with breaking points precisely at points y_i. Thus, it attains its minimum value at one of the points y_i. Depending on the actual values of y_i, the point c point might be determined uniquely, or it might lie within a closed interval delimited by two neighbouring y_is (for ... 2 NOTICE This is an answer to the original question. While I was writing the answer the OP changed the condition a_n/b_n\to\infty to b_n/a_n\to\infty. Take b_n=a_n^2. Since \sum a_n converges, a_n\to0, and a_n/b_n=1/a_n\to\infty. Also, a_n is bounded. Let A be an upper bound. Then 0\le b_b\le A\,a_n, so that \sum b_n converges. 2 i think the question has been well-answered, but would just like to draw your attention to the fact that there is no mystery here. let us leave aside the distraction of the partial derivative. suppose suitable differentiability of the functions involved, and that you wish to find:$$\frac{d (\phi \circ f)(x)}{d\phi(x)} $$then$$ d(\phi \circ f)(x) = ...

2

Hint: The function you need to integrate is $$(3t - 8\mathrm s)^{-2}\mathrm m = \dfrac{1\mathrm m}{(3t - 8\mathrm s)^2}$$ If you don't immediately see how we can apply the power rule in integration, you can use $u$-substitution? Let $u = 3t - 8\mathrm s\implies du = 3\,dt \iff \frac 13 du = dt.\;$ Then your integrand becomes $$\dfrac {1\mathrm m}3 ... 2 Substitute x = R\tan \theta then dx = R \sec^2 \theta d\theta and x^2+R^2 = R^2(\tan^2\theta + 1) = R^2 \sec^2 \theta. Note as well that x = R\tan \theta implies that \sin \theta = \dfrac{x}{\sqrt{x^2+R^2}}.$$ \int \frac{Rdx}{\left(x^2+R^2\right)^{3/2}} = \int \frac{R^2 \sec^2 \theta d\theta}{\left(R^2 \sec^2 \theta\right)^{3/2}} = \int ...

2

$$f(x,y) = 3x - 4x^3 + 12 xy \Rightarrow \frac{\partial f}{\partial x} = 3 - 12x^2 + 12y, \frac{\partial f}{\partial y} = 12x$$ You have to pose derivatives equal to $0$: $$\frac{\partial f}{\partial y} = 12x = 0 \Rightarrow x = 0$$ $$\frac{\partial f}{\partial x} = 3 - 12x^2 + 12y = 3 + 12y = 0 \Rightarrow y = -\frac{1}{4}$$ The point $A = (0, ... 1 If$x^*$is an interior point of$\Omega$, then every direction$d \in \mathbb{R}^n$is a feasible direction from$x^*$, i.e., for any fixed$d$,$x^* + \alpha d \in \Omega$for any sufficiently small values of$|\alpha|$. Defining$g(\alpha)$as was done above, we again Taylor expand$g(\alpha)$g(\alpha) = g(0) + g'(0)\alpha + ... 1$\sin^2{A}+\cos^2{A}=1$implies$\tan^2{A}+1=\sec^2{A}$, so we let$x=R\tan{A}$and dx=R\sec^2{A}dA$, and (x^2+R^2)^{3/2}=R^3\sec^3{A}$so the integral becomes $$\int\frac{R^2\sec^2{A}}{R^3\sec^3{A}}.$$ Which is$\frac{1}{R}\int \cos{A}dA$. So it integrates as, $$\frac{\sin{A}}{R}.$$ Drawing a right triangle with tangent equal to$\frac{x}{R}$we see it's ... 1 HINT: Use Lagrange multipler with KKT conditions. You'll have: $$F(x,y,\lambda,\lambda_1,\lambda_2) = 3x - 4x^3 + 12xy - \lambda(x+y-1) - \lambda_1x - \lambda_2y$$ Now take partial derivatives and because those lambda things have to be zero, set them to zero. We know that if a product is zero then one of the multiples is zero. So check every possible ... 1 You need to be sure to use a unit normal. Since$\text{curl }F\cdot n$is constant, you'll just get a constant times the area of the disk. No parametrization or explicit integration needed. By the way, it's easier to get the normal directly from the linear equation$a\cdot x=0$: We get$a=(1,2,2)$, and so$n=(1,2,2)/3$. 1 The curve$z=\log_b(xR+1)$has$z(0)=0$and$z(C)=k$provided$R$is defined as $$R=\frac{b^k-1}{C}.\tag{1}$$ From a comment it appears you want$B$to denote the value of$x$at which$z(x)=1$, and since$z$is defined as a log base$b$this means that at$x=B$the input of the log should be$b$[since$\log_b(b)=1$]. This gives$BR+1=b$or ... 1 I'm not sure what techniques you're expected to use, but a very basic way of showing it is to differentiate the two equations and solve for$\dfrac{dy}{dx}$in both. This gives you the equations of the tangent lines at all points. Orthogonal lines have slopes that are the negative reciprocal of each other and that's precisely what we get. Differentiating ... 1 No, it is not: The condition $$s(f, f) = 0 \implies f = 0$$ does not hold. Try coming up with a specific example to show this; a piecewise defined function could work well here. In fact, choosing any function$f$which is negative on$[0, 1/2]$and positive on$[1/2, 1]$will give an example of $$s(f, f) < 0$$ for the integrand will always be ... 1 To find the velocity, you have to find the integral of the acceleration function.$v(t)=\int a(t) dt=\int (3t-8)^{-2} dt=-\frac{1}{3(3t-8)}+C$We know that$v(0) = 2$, and so we know that$C$will equal:$2=\frac{1}{24}+C; C=\frac{47}{24}v(t)$is then:$v(t) = -\frac{1}{3(3t-8)}+\frac{47}{24}$We want$v(6)$so just plug 6 in for t and you will get: ... 1 Hint: In the left-hand side,$x_1$,$x_2$,...$x_n$can take arbitrary values between$a$and$t$, in the right-hand side, you have$t\geq x_1\geq x_2\geq\dots\geq x_n\geq a$... and a factor$n!$which is the number or permutations of$n$elements. Edit: you should first write ... 1 The only condition we need is$f$Lebesgue-integrable. I assume without loss of generality that$a=0$and$t=1$. Define$g(x_1,\dots,x_n):=f(x_1)\dots f(x_n)$,$I:=[0,1]^n$and$S:=\{(t_1,\dots,t_n)\in I,0\lt t_1\lt\dots\lt t_n\lt 1\}$. Then $$I':=\{(t_i)_{i=1}^n,i\neq j\Rightarrow t_i\lt t_j\}=\bigsqcup_{\sigma\in\mathcal ... 1$$ \frac{\partial\log\mu(x)}{\partial\log x_j}\cdot\frac{\partial\log x_j}{\partial x_j}=\frac{\partial \log \mu(x)}{\partial x_j}=\frac{\partial\mu(x)}{\partial x_j}\frac{1}{\mu(x)} $$Since \frac{\partial \log x_j}{\partial x_j}=\frac{1}{x_j}, we obtain:$$ \frac{\partial\log\mu(x)}{\partial\log x_j}=\frac{\partial\mu(x)}{\partial x_j}\frac{x_j}{\mu(x)} ... 1 The first step in any problem is to draw a picture: We are told that$\frac{d\theta}{dt} = 10\pi \frac{\text{rad}}{\text{min}}$, and we wish to find$\frac{dx}{dt}$when$x = 50$. We can write an equation relating$x$and$\theta$using tangent: $$\tan{\theta} = \frac{x}{100\,\text{yds}} \tag1$$ Differentiating$(1)$with respect to time: ... 1 Letting$f(x)$be the left hand side of the equation, then imagine the graph of$y=f(x)$. We can get the followings : $$\lim_{x\to \pm\infty}f(x)=0, \lim_{x\to\lambda_i\pm0}f(x)=\pm\infty.$$ With the continuity of the graph except$x=\lambda_i$, this leads what you desire. Do you see the monotonicity of each part? The answer$X$will be in$\lambda_1\lt ...

1

Just examine the signs of the function on the left hand side (call it $f$) for all intervals where it is continuous. For $x < \lambda_1$, $f$ is always negative. So there are no roots here. For $\lambda_1 < x < \lambda_2$, $f(x) \to \infty$ as $x \to {\lambda_1}^+$ but $f(x) \to -\infty$ as $x \to {\lambda_2}^-$. So there is a root here. Also, $f$ ...

1

We want, $$\left(\int_{0}^{4} \sqrt{x}\,\,dx-\int_{0}^{4}\dfrac12x\,\,dx\right)+\left(\int_{4}^{16}\dfrac12x\,\,dx-\int_{4}^{16}\sqrt{x}\,\,dx\right)$$ where the first term two terms give the area from $0$ to $4$ and the second term gives it from $4$ to $16$. The expression becomes: ...

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