Is there a way to find the derivative of an equation where I cannot separate the independent variable from the dependent variable. For example, what is $dy/dx$ given  \begin{equation*} \cos(y) = \dfrac{x}{x+y} \end{equation*}
121368
I have to use implicit differentiation to find $\frac{dy}{dx}$ given:  $$x^2 \cos(y) + \sin(2y) = xy$$  I don't even know where to begin, I missed the class where we went over implicit differentiation, and because of that, I am completely stuck.   Thank you everyone.  Edit: I don't know how to make the equation look all nice and whatnot, so sorry about that
22772
Given $f = f(u(t), t)$, is it possible to find the following derivative? ($f$,$u \in \mathbb{R}^n$, $t \in \mathbb{R}$)  $$ \frac{\partial f}{\partial||u||^2_2} $$  I am aware of the following derivatives:  $$ \frac{\partial||x||^2_2}{\partial x} = \frac{\partial ||x^Tx||_2}{\partial x} = 2x $$  So I can invert it to give me (elementwise division in the RHS)  $$ \frac{\partial x}{\partial||x||^2_2} = \frac{1}{2x} $$  This step does feel like abuse of notation I find it hard to convince myself of its validity.  If it's right though, I could just apply the chain rule to find  $$ \frac{\partial f}{\partial u} \frac{\partial u}{\partial||u||^2_2} $$  However, knowing that both $f$ and $u$ are a function of $t$, I am somehow doubting my own reasoning. Or does that just mean that the derivative will still be a function of $t$ and I have nothing to worry about? Can anyone shed some light?  UPDATE: I'm now doubting the existence of  $$ \frac{\partial u}{\partial ||u||^2_2} $$  because $u = u(||u||_2^2)$ is not injective. Is the inversion step I did above legitimate? Is there another possible approach?
39920
$$f(x,y) = \begin{cases} \displaystyle \frac{xy(x^2-y^2)}{x^2+y^2} &amp; \text{if } (x,y) \neq (0,0), \\ 0 &amp; \text{if } (x,y) = (0,0). \end{cases}$$ I tried finding both mixed partial derivatives but they ended up being the same for that function. I must be failing to take into account something dealing with the fact that it is piece-wise. I still need to show the mixed partial derivatives exist. How can I do all of this?
89551
I looked at the graph of $f(x,y)=(xy)^{2/3}$ at the origin, and no matter how far I zoom in it has four corners along $x=\pm y$, so it doesn't look like it is differentiable as $z=0$ doesn't look like a tangent plane.  I also found that its partial derivatives are not continuous at $(0,0)$, e.g.  $$ \lim_{(ky^2,y) \to (0,0)} \frac{\partial f}{\partial x} = \lim_{(ky^2,y) \to (0,0)} \frac{2}{3}x^{-\frac{1}{3}}y^{\frac{2}{3}} = \frac{2}{3}k^{-\frac{1}{3}} $$ (although $\frac{\partial f}{\partial x}(0,0) = \frac{\partial f}{\partial y}(0,0) = 0$)  I also tried to prove/disprove its differentiability by definition, and for it to be differentiable I need $$ \lim_{(x,y) \to (0,0)} \frac{f(x,y)-0}{\sqrt{x^2+y^2}} =0 $$ but then I got stuck. Am I even on the right track?  [UPDATE] I need to do the same thing for $f(x,y)=(xy)^{\frac{1}{3}}$... I think the answer is that $(xy)^{2/3}$ is differentiable but $(xy)^{1/3}$ is not, according to my professor but I'm not sure
101346
The definition of a partial derivative is the "derivative of a multi-variable function relative to a single variable when all other variables are held constant".  But isn't the regular derivative (for one-variable functions) just a trivial case of this, where there are no other variables to hold constant? Why do we need the separate notation for partial derivatives (that is, writing $\displaystyle \frac{\partial f}{\partial x}$ rather than just $\displaystyle \frac{df}{dx}$)?
7365
In the book Thomas's Calculus (11th edition) it is mentioned (Section 3.8 pg 225) that the derivative $dy/dx$ is not a ratio. Couldn't it be interpreted as a ratio, because according to the formula $dy = f'(x)dx$ we are able to plug in values for $dx$ and calculate a $dy$ (differential). Then if we rearrange we get $dy/dx$ which could be seen as a ratio.   I wonder if the author says this because $dx$ is an independent variable, and $dy$ is a dependent variable, for $dy/dx$ to be a ratio both variables need to be independent.. maybe?
17531
I've recently been learning about differential equations, and my teacher has been giving some particularly difficult examples out for those of us who finish early. He gave us the following differential equation: $$x=y\frac{dy}{dx}-\left(\frac{dy}{dx}\right)^{2}$$ With the hint "Differentiate with respect to $y$, then let $\frac{dy}{dx}=p$". I've solved it after the lesson and would like someone to check whether or not I've done so correctly.    Following the hint, $$\frac{dx}{dy}=\frac{1}{p}=p+y\frac{dp}{dy}-2p\frac{dp}{dy}$$ Which rearranges to $$p-\frac{1}{p}+y\frac{dp}{dy}=2p\frac{dp}{dy}$$ Multiplying through by $\frac{dy}{dp}$, $$\left(p-\frac{1}{p}\right)\frac{dy}{dp}+y=2p$$ Rearranging into the form $\frac{dy}{dp}+f(p)y=g(p)$, and writing $p-\frac{1}{p}=\frac{p^{2}-1}{p}$ $$\frac{dy}{dp}+\frac{p}{p^{2}-1}y=\frac{2p^{2}}{p^{2}-1}$$ Our integrating factor is $$e^{\int\frac{p}{p^{2}-1}dp}=e^{\frac{1}{2}\ln(p^{2}-1)}=\sqrt{p^{2}-1}$$ So we have $$\sqrt{p^{2}-1}\frac{dy}{dp}+\frac{p}{\sqrt{p^{2}-1}}y=\frac{2p^{2}}{\sqrt{p^{2}-1}}$$ The left hand side is, by design, $\frac{d}{dp}(y\sqrt{p^{2}-1})$, and the right hand side can be integrated as follows: $$\int \frac{2p^{2}}{\sqrt{p^{2}-1}}dp=\int\frac{2(p^{2}-1)+2}{\sqrt{p^{2}-1}}dp=\int 2\sqrt{p^{2}-1}dp+\int \frac{2}{\sqrt{p^{2}-1}} dp$$ These can both be solved using the substitution $p=\cosh(u)$, and gives $$y\sqrt{p^{2}-1}=p\sqrt{p^{2}+1}-\cosh^{-1}(p)+\alpha$$ Where $\alpha$ is an arbitrary constant. Dividing, we finally have $$y=p+\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}$$  Here I was stuck for a long time, until I realised that we also have the equation we began with: $x=py-p^{2}$, which we can solve as a quadratic in $p$ to also get $$p=\frac{y\pm \sqrt{y^{2}-4x}}{2} \implies p^{2}=\frac{y^{2}+y^{2}-4x \pm 2y\sqrt{y^{2}-4x}}{4}$$  Thus, $$y=\frac{y\pm \sqrt{y^{2}-4x}}{2}+\frac{\cosh^{-1}\left(\frac{y\pm \sqrt{y^{2}-4x}}{2}\right)+\alpha}{\sqrt{\frac{y^{2}-2x \pm y\sqrt{y^{2}-4x}}{2}-1}}$$  This is as good as I can do - we have a relation between $y$ and $x$ - am I done? Is this the correct answer? Is there an easier way?     Thank you for your time.    Following the suggestion by @Valentin $$\frac{dy}{dp}=1+\frac{\frac{1}{\sqrt{p^{2}-1}}\sqrt{p^{2}-1}-p\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}}{p^{2}-1}=1+p\frac{1-\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}}{p^{2}-1}$$ Hence, $$dx=\frac{dy}{p}=dp\left(\frac{1}{p}+\frac{1-\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}}{p^{2}-1}\right)$$
90387
Use the definition of the derivative to find $f'(x)$ for $f(x)=\sqrt{x-2}$. I don't even know where to start with this. I have done $\sqrt{x-2}= (x-2)^.5$.  Is this correct?
104034
When \begin{eqnarray*} y'[x] &amp;=&amp; \frac{r y[x] }{ x } \; \;\text{and} \;\; y[1] = a \\ y[x] &amp;=&amp; a x^r \end{eqnarray*} Use derivative formulas to explain the output.  What I have so far:  We know that  \begin{eqnarray*} y'[x] = r y[x] = k e^{rx} \end{eqnarray*} So we can rewrite as  \begin{eqnarray*} y[x] = \frac{k e^{rx} } { x} \end{eqnarray*} To get the value of k: \begin{eqnarray*} a = \frac{k e^{r} }{ 1} \\ k = \frac{a}{e^r} \end{eqnarray*} So now we have \begin{eqnarray*} \frac{(a/(e^r)) e^{rx} } { x} \end{eqnarray*} But when I plug this into a derivative calculator I get a solution much different than $y[x] = a x^r$. Perhaps I am approaching this wrong. What is the correct way to approach this problem using derivative formulas?
53224
I am a 12th grade student, and I am afraid that in realistic terms this question might not even make sense because of the infinities that have to be dealt with. However, in my attempt to calculate it's derivative, I did the following:  $$y=x^{x^{x^{.^{.^.}}}}$$  $$\ln(y)=\ln(x)x^{x^{x^{.^{.^.}}}}$$  $$\ln(y)=y\ln(x)$$  after taking derivative with respect to x on both sides, I obtained the following:  $$\frac{\mathrm dy}{\mathrm dx}=\frac{y^2}{x(1-\ln(y))}$$  I am fairly certain that the calculations up to this point are valid. However, to further continue, I analyzed the nature of y in different domains and obtained the values, by observation(that is, by observing how $x$, $x^x$, $x^{x^x}$, $x^{x^{x^x}}$, and so on would behave to draw a conclusion):   for $x&lt;1,$ $y$ becomes 0 for $x=1,$ $y$ becomes 1 for $x&gt;1,$ $y$ becomes infinite   These 3 points is where the first problem lies. Are these true? If so, how do we reach to the conclusion?  Secondly, considering this to be true, I get the derivative at:   $x&lt;1$ to be 0. $x=1$ to be 1. for $x&gt;1$, I took $x=2.$ then the derivative $\frac{\mathrm dy}{\mathrm dx} = y^2/[2(1-ln(y))]$ (replacing $x$ by $2$). Now, I applied L hospital's rule to get the value of the expression to be negative infinity.   This is the second problem. I have used L Hospital's rule, but limits were not concerned. Is this method valid? If not, how would we calculate it?
110927
I have:  $\dfrac {dx} {dt}$=$-x+y$  $\dfrac {dy}{dt}$=$-x-y$  and I am trying to find $x(t)$ and $y(t)$ given that $x(0)=0$ and $y(0)=1$  I know to do this I need to decouple the equations so that I only have to deal with one variable but the decoupling is what I am having trouble with  Do I set them equal to each other and then just move like terms to separate sides getting two different equations and then integrate?
112183
Let $f =f(x(t),y(t))$  Then the total derivative of f with respect to t is given by:  $$\frac {}{} = \frac {}{} + \frac {}{}$$  Say still we have  $f=f(x(t),y(t))$. What is then $\frac {\partial^2 f}{ \partial t^2} $? Is it just $$ \frac {\partial f}{ \partial x} \frac {d^2x}{dt^2}+ \frac {\partial f }{\partial y} \frac {d^2y}{dt^2} $$ ? (Just by applying chain rule again?)  Also, this seems to be related to the Jacobian  J=J(f). Is this correct? If so, can we define $\frac{\partial^2 f}{ \partial t^2}$ as $JJ^T$ , where $J^T$ is the transpose of $J$?
103517
Given a matrix $A$ and column vector $x$, what is the derivative of $Ax$ with respect to $x^T$ i.e. $\frac{d(Ax)}{d(x^T)}$, where $x^T$ is the transpose of $x$? Side note - my goal is to get the known derivative formula $\frac{d(x^TAx)}{dx} = x^T(A^T + A)$ from the above rule and the chain rule.
106400
How would I find the derivative of a unit step function? I understand that the unit impulse function will be used but I'm not sure how to use it.  I am trying to find the derivative of this:  $v(t) = u(t+1) - 2u(t) + u(t-1)$  $u(t) = 0$  when $t &lt; 0$  $u(t) = 1$ when $t &gt; 0$  The relationship between unit step function and impulse function:  &delta;(n) = u(n) - u(n-1)  $ δ(t)=du(t)/dt $
42960
One problem I have found when learning calculus is that there are many different ways to denote the derivative. If $y=f(x)=x^2$, then we could write \begin{align} f'(x)&amp;=2x \\ y'&amp;=2x \\ \frac{df}{dx}(x)&amp;=2x \\ \frac{df(x)}{dx}&amp;=2x \\ \frac{d}{dx}f(x)&amp;=2x \\ \frac{dy}{dx}&amp;=2x \end{align} And this is just Lagrange and Leibniz's notations alone. What I find troubling is that they all seem to be suggesting subtly different things about what the derivative actually is. Is it a function, a limit of a quotient, or both? In the interests of keeping my post brief, I'll focus my attention on $f'(x)=2x$ and $\frac{dy}{dx}=2x$, as these seem to be the most common notations.  $$ f'(x)=2x $$  It does make sense to think of the derivative as the gradient function: $$ f'\colon x\mapsto\lim_{\Delta x \to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} $$ In this case the limit expression is equal to $2x$, and so we can write $$ f' \colon x \mapsto 2x $$ However, this notation seems a little counter-intuitive when we consider what it means to differentiate a function with respect to a variable other than $x$. If I ask what is the derivative of $f(x)$ with respect to $\frac{x}{2}$, does this question make sense? Is it simply $f'(\frac{x}{2})$? Or do we have to express $x^2$ in terms of $\frac{x}{2}$? And how can we can express this derivative using Lagrange's notation?  $$ \frac{dy}{dx}=2x $$  There are many things which are nice about Leibniz's notation, including the fact that it is explicit which variable you are differentiating with respect to. However, in this case, it is unclear whether we are talking about a function, or something else entirely. There are other issues. Some people say they dislike the Leibniz formulation of the chain rule $$ \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx} $$ saying that they find it to be inaccurate. I don't really understand why this is the case. Could someone please elaborate?
92873
Please guide me:  $y' + ay +b = 0$ ($a$ not zero) is supposed to be separable and has solution  $y = ce^{-ax} - \frac ba$  Here is my start to this problem:  $\frac{dy}{dx} + ay = -b$ is as far as I can go with this. How should I go about separating $x$ and
99618
While I do know that $\frac{dy}{dx}$ isn't a fraction and shouldn't be treated as such, in many situations, doing things like multiplying both sides by $dx$ and integrating, cancelling terms, doing things like $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$ works out just fine.  So I wanted to know: Are there any particular cases (in single-variable calculus) we have to look out for, where treating $\frac{dy}{dx}$ as a fraction gives incorrect answers, in particular, at an introductory level?  Note: Please provide specific instances and examples where treating $\frac{dy}{dx}$ as a fraction fails