7

Consider the following function. It's continuous on $[0,1]$, differentiable on $(0,1)$ but neither at $0$ nor $1$. Yet Rolle's theorem still applies and guarantees there is an $x$ in $(0,1)$ such that $f'(x)=0$. It's for this kind of situation that it's fundamental to distinguish between continuity on $[a,b]$ and differentiability on $(a,b)$. Had Roll's ...


6

Chain rule is $$ \left(\frac d{dx} (f\circ g)\right) (x) = \left(\frac{df}{dx}\circ g\right)(x)\frac{dg}{dx}(x),$$ or using the $f'$ notation: $$ (f\circ g)'(x) = (f'\circ g)(x) g'(x).$$ When we apply this with $g(x) = -x$, we see that the correct rule is $$\frac{d}{dx} (f(-x)) = -f'(-x).$$ (This is what your derivation says, even if you stated it wrong in ...


6

I'd use $y = xv$ to homogenize the right side (i.e. $y^2+x^2 = x^2(1+v^2))$ $$y' = v+xv' = \frac{xv+x^2v^2+x^2}{x} = v + x(1+v^2)$$ $$\implies v' = 1+v^2$$ which has the solution $v = \tan(x+C)$ by separation of variables


5

For $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the equation of the tangent line at $(a\cos t, b \sin t)$ is given by $$\frac{\cos t}ax+\frac{\sin t}by=1$$ which intersects the axes at $\frac a{\cos t}$ and $\frac b{\sin t}$, respectively. Then, the area of the triangle is $$Area = \frac12 \frac a{\cos t}\frac b{\sin t}=\frac{ab }{\sin2t}\ge ab $$ Here, $a=\frac1{...


5

The answer is yes. In order to avoid confusing fractions and differentials everywhere, I will write this in purely operator form to emphasize how simple this is. We can denote differentiation as a map between sets of differentiable/integrable functions, i.e $$\mathcal{D}:\mathscr{D}\to\mathscr{I}~~,~~\mathcal{D}:f\mapsto \mathcal{D}(f)$$ Integration can be ...


5

Your proof is not correct as $\delta$ depends on $x$ in contradiction with what is requested. You can proceed as follows. Choose $\epsilon \gt 0$. According to Heine–Cantor theorem it exists $\delta \gt 0$ such that $\left\vert f^\prime(t) - f^\prime(x) \right\vert \lt \epsilon$ for $\vert t - x \vert \lt \delta$. To get the conclusion, pickup $x,y$ with $\...


5

Take polar coordinates $$\lim_{r\to 0} \frac{r^4\cos^2\theta \sin^2\theta}{r^2(\cos^2\theta + \cos\theta\sin\theta + \sin^2\theta)} = \lim_{r\to 0} r^2 \frac{\cos^2\theta \sin^2\theta}{\cos^2\theta + \cos\theta\sin\theta + \sin^2\theta}$$ For this limit to exist, we have to able to approach the same value for every direction. Does it happens in this case?


5

The function $f$ is not differentiable at $(0,0)$. If it was, then, since $f_x(0,0)=f_y(0,0)=0$, then $f'(0,0)$ would be the null function. In other words,$$\lim_{(x,y)\to(0,0)}\frac{x^2y}{(x^2+y^2)^{3/2}}=0.$$But if $x=y>0$, $\frac{x^2y}{(x^2+y^2)^{3/2}}=\frac1{2\sqrt2}$.


5

Hint: Let $I=[-1,1]$. $f(t)=t^2$, $g(t)=t\left|t\right|$.


5

$$\frac{du}{u}=-\ln(u(t))dt$$ You cannot integrate $\int-\ln(u(t))dt$ directly. The mistake is : You integrate $\quad \int-\ln(u(t))du= -u\ln(u)+u\quad$ but this is not $\quad \int -\ln(u(t))dt$. Back to the original ODE : $$\frac{du}{dt}=u\ln(u)$$ This is a separable ODE thus $$\frac{du}{u\ln(u)}=dt$$ $$\int\frac{du}{u\ln(u)}=\int dt$$ $$\ln(\ln(u))=t+c$$ $$...


4

No to both. $f(x) = x$ is differentiable at $x = 0$, but $|x|$ is not. The sign function $$f(x) = \begin{cases} 1 & \text{if } x \geq 0 \\ -1 & \text{if } x < 0 \end{cases}$$ is not even continuous at $0$, but $|f(x)| = 1$ for all $x$, therefore $|f|$ is differentiable everywhere.


4

You made a mistake in your final integral. The upper end-point's dependence on $x$ means you should get $$\frac{{\rm d}}{{\rm d} x} \int_0^x u(t) \ {\rm d} t = u(x)$$ In general, $$\frac{{\rm d}}{{\rm d} x} \int_0^x g(x,t) \ {\rm d} t = \int_0^x \frac{\partial}{\partial x} g(x,t) \ {\rm d} t + \boxed{g(x,x)}$$ But throughout the question, $g(x,t) \propto (...


4

No, nothing can be said about the sign of $f''$ in the vicinity of an isolated minimum. An example is the function $f$ defined by $$ f(x) = e^{-1/x^2}\left(3 + 2 \sin(\frac {1}{x^2})\right) $$ for $x \ne 0$ and $f(0) = 0$. $f$ is infinitely often differentiable and has an absolute minimum at $x=0$. For $x \ne 0$ is $$ f''(x) = \frac{2e^{-1/x^2}}{x^6 } \...


4

You can approach this question by first solving for $y$ to get $$y=\frac{2x}{4-lnx}$$ From here, use the quotient rule to get $$\frac{dy}{dx}=\frac{2(4-lnx)-(-\frac{1}{x}(2x)}{(4-lnx)^2}=\frac{8-2lnx+2}{(4-lnx)^2}=\frac{10-2lnx}{(4-lnx)^2}$$ Substituing $x=1$ we get $$\frac{dy}{dx}=\frac{10-2ln(1)}{(4-ln(1))^2}=\frac{10}{16}=\frac{5}{8}$$ Therefore, the ...


4

Note that $f(x) $ must have even degree and a positive leading coefficient (unless it is zero, in which case it's trivial) and thus so must $g(x)$. Thus, $g$ has a global minimum, say at $y$. From the definition of $g$, $g'(y)=0$ then implies that $g(y) = f(y) \ge 0$.


4

Some assumptins are missing. A counter-example is $f(x)=x-1$ and $g(x)=2x$ in which case $h$ is increasing.


4

$$\frac{d}{d x}\left(\int_0^{y(x)} e^{t^2} \, dt+\int_0^x \sin (t) \, dt\right)=0$$ For the Fundamental theorem of Calculus $$e^{y(x)^2} y'(x)+\sin (x)=0$$ solving wrt $y'$ $$y'(x)=-e^{-y(x)^2} \sin (x)$$


4

No, the given conditions do not imply that $f(e^x)$ is convex in some interval $(-\delta, 0]$. We will give a concrete counterexample below. Preliminaries/motivation If $f$ is differentiable then $g(u) = f(e^u)$ is convex iff $h(x) = xf'(x)$ is increasing. We have that $h$ is negative and $h(1) = 0$. Of course that does not imply that $h$ is increasing near $...


4

Since the partial derivatives exist and are continuous at every point, then the function is differentiable at every point.


4

Hint: Rewrite the DE as $$\frac{dx}{dy}-\frac{x}{y}=y$$ which is easy by integrating factor method


3

For the first you should use that the composition of continuous functions is continuous to conclude the exercise. For the second note that $x^2$ is a counterexample, as $\sqrt{x^2}=|x|$ which is clearly non differentiable.


3

We have a function $f(x)$ that is differentiable on $(-\infty,\infty)$ and with range $[0,\infty)$, which is the most general form of your function. Now, take the derivative of $\sqrt{f(x)}$. By the chain rule, we get that the derivative of $\sqrt{f(x)}$ is $\frac{f'(x)}{2\sqrt{f(x)}}$. We already know that $f(x)$ is differentiable on $(-\infty,\infty)$, so ...


3

We have $f(x)=\dfrac{e^{-2x}}{\sqrt x}$. Use Quotient rule as you suggested in your answer too: $$f'(x)=\frac{-2e^{-2x}\sqrt x -\frac{1}{2\sqrt x}\times e^{-2x}}{x}$$ Now multiply the fraction by $\frac{2\sqrt x}{2\sqrt x}$ to get rid of $\frac{-1}{2\sqrt x}$ in numerator: $$\frac{-2e^{-2x}\sqrt x -\frac{1}{2\sqrt x}\times e^{-2x}}{x}\times\frac{2\sqrt x}{2\...


3

If $A = (a_{ij})_{ij}$ then $\operatorname{div}(A)_i$ (that is, the $i$-th component of the vector $\operatorname{div}(A)$) is given by $$\sum_{j}\partial_j a_{ij}.$$ If $A = (\nabla u)^t$, that is, if $A$ is the transpose of $\nabla u$, then $a_{ij} = \partial_i u_j$ and therefore $$\operatorname{div}((\nabla u)^t)_i = \sum_{j}\partial_j\partial_iu_j.$$ On ...


3

It's sufficient to consider the case $\omega=1$: if we let $$ \color{DarkBlue}{f_n(x):=e^{\sqrt{x}}\sqrt{x}\frac{d^n}{dx^n}\frac{e^{-\sqrt{x}}}{\sqrt{x}}}, $$ then the needed expression is $$ \frac{d^n}{dx^n}\frac{e^{-|\omega|\sqrt{x}}}{\sqrt{x}}=|\omega|^n\frac{e^{-|\omega|\sqrt{x}}}{\sqrt{x}}f_n(\omega^2 x). $$ To compute $f_n(x)$, we consider (for $x>0$...


3

First, judging from your drawing, you probably mean horizontal strips not vertical strips. That aside, this problem is an application of IVT (intermediate value theorem), MVT (mean value theorem) and certain other facts leveraging the continuity of $f$ (e.g. preimages of closed sets are closed). As $f(x_0) < a < f(x_1)$, by IVT there is some $x_0 < ...


3

$$2+y' \ln x + y \cdot \frac1x = 4 y'$$ now plug in the point and solve for $y'$ You might want to revise "product rule" and "implicit differentiation"


3

Your curve is given by the equation $$4y = 2x + y\ln(x)$$ Which means $$y = \frac{2x}{4 - \ln(x)}$$ At this point, the equation of the tangent line at this curve is given by $$y_t = y(x_p) + y'(x_p)\cdot(x - x_p)$$ Where $x_p$ is the $x-$ coordinate of your point $P$ which means $x_p = 1$. $y(x_p)$ is your curve evaluated at $x = x_p = 1$ and $y'(x_p)$ is ...


3

You can either rearrange to get $y$ in terms of $x$ or use implicit differentiation. I prefer to use rearrangement. If you would like me to use implicit differentiation I will be more than happy to oblige. We have $$\begin{align} 2x+y\ln x&=4y\\ \implies4y-y\ln x&=2x\\ \implies y(4-\ln x)&=2x\\ \implies y&=\frac{2x}{4-\ln x} \end{align}$$ ...


Only top voted, non community-wiki answers of a minimum length are eligible