What is the derivative of $e}^{\mathrm{cos}x}+{e}^{\mathrm{sin}y}=\frac{1}{4$

Amaris Orr
2022-09-23
Answered

What is the derivative of $e}^{\mathrm{cos}x}+{e}^{\mathrm{sin}y}=\frac{1}{4$

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brodireo1

Answered 2022-09-24
Author has **12** answers

What to you mean exactly? You can calculate the derivative of a function in the form $y=f\left(x\right)$ (or, in your case, $z=f(x,y)$, but what do you mean with the derivative of something like $f(x,y)=c$? You only want to know which the derivative of $e}^{\mathrm{cos}\left(x\right)}+{e}^{\mathrm{sin}\left(x\right)$ is?

asked 2022-11-23

I need to find $\frac{dy}{dx}$:

$\frac{d}{dx}{e}^{y}\mathrm{cos}x=1+\mathrm{sin}(xy)$

using implicit differentiation.

So far I've gotten to this point which I don't think I'm doing right because then I get stuck:

${e}^{y}(-\mathrm{sin}x)+(\mathrm{cos}x)({e}^{y})(\frac{dy}{dx})=\mathrm{cos}(xy)(x)(\frac{dy}{dx})+y(1)$

Am I right? Am I wrong? If I'm right I definitely have no clue where to go...and if I'm wrong...then again I have no idea what to do.

$\frac{d}{dx}{e}^{y}\mathrm{cos}x=1+\mathrm{sin}(xy)$

using implicit differentiation.

So far I've gotten to this point which I don't think I'm doing right because then I get stuck:

${e}^{y}(-\mathrm{sin}x)+(\mathrm{cos}x)({e}^{y})(\frac{dy}{dx})=\mathrm{cos}(xy)(x)(\frac{dy}{dx})+y(1)$

Am I right? Am I wrong? If I'm right I definitely have no clue where to go...and if I'm wrong...then again I have no idea what to do.

asked 2022-08-19

Let $z=z(x,y)$ be defined implicitly by $F(x,y,z(x,y))=0$, where $F$ is a given function of three variables.

Prove that if $z(x,y)$ and $F$ are differentiable, then

$\frac{dz}{dx}=-\frac{\frac{dF}{dx}}{\frac{dF}{dz}}$

if $dF/dz\ne 0$

I am kind of confused with $z=z(x,y)$. Should I differentiate $dz$/$dx$ and $dz$/$dy$, but there's nothing inside $x$ and $y$ so what should I differentiate $dx$ and $dy$ with then? I've been told that this is something related to implicit differentiation. In some other examples, they do use differentiate $F$ i.e. $dF$ but in most examples, I only see that they are differentiating the variables inside it.

Prove that if $z(x,y)$ and $F$ are differentiable, then

$\frac{dz}{dx}=-\frac{\frac{dF}{dx}}{\frac{dF}{dz}}$

if $dF/dz\ne 0$

I am kind of confused with $z=z(x,y)$. Should I differentiate $dz$/$dx$ and $dz$/$dy$, but there's nothing inside $x$ and $y$ so what should I differentiate $dx$ and $dy$ with then? I've been told that this is something related to implicit differentiation. In some other examples, they do use differentiate $F$ i.e. $dF$ but in most examples, I only see that they are differentiating the variables inside it.

asked 2022-08-26

Given:

${e}^{7z}=xyz$

The task is to compute the partial derivatives dz/dx and dz/dy using implicit differentiation. My solution is now as follows:

$(dz/dx)7{e}^{7z}=yz+(dz/dx)xy$

And so,

$(dz/dx)(7{e}^{7z}-xy)=yz$

Therefore correct answer is,

$\frac{dz}{dx}=\frac{yz}{(7{e}^{7z}-xy)}$

${e}^{7z}=xyz$

The task is to compute the partial derivatives dz/dx and dz/dy using implicit differentiation. My solution is now as follows:

$(dz/dx)7{e}^{7z}=yz+(dz/dx)xy$

And so,

$(dz/dx)(7{e}^{7z}-xy)=yz$

Therefore correct answer is,

$\frac{dz}{dx}=\frac{yz}{(7{e}^{7z}-xy)}$

asked 2022-10-11

Find $\frac{dy}{dx}$ by implicit differentiation.

$\mathrm{tan}(x+y)=x$

So far, I got to

${\mathrm{sec}}^{2}(x+y)(1+\frac{dy}{dx})=1$

but then im lost.. can someone please help and explain? I would really appreciate it!!

$\mathrm{tan}(x+y)=x$

So far, I got to

${\mathrm{sec}}^{2}(x+y)(1+\frac{dy}{dx})=1$

but then im lost.. can someone please help and explain? I would really appreciate it!!

asked 2022-09-23

Consider the implicit function

$(w(x)+1){e}^{w(x)}=x.$

I need to approximate $w(1.1)$ using the fact that $w(1)=0$.

Could you give me any hints?

$(w(x)+1){e}^{w(x)}=x.$

I need to approximate $w(1.1)$ using the fact that $w(1)=0$.

Could you give me any hints?

asked 2022-08-25

Find $\frac{dy}{dx}$ by implicit differentiation

${x}^{2}-4xy+{y}^{2}=4$

I know to take the derivatives of both sides, which would be:

$\frac{d}{dx}[{x}^{2}-4xy+{y}^{2}]=0$

I'm not sure if I did it right, but I then got:

$2x-4\cdot (x\frac{dy}{dx})+y+2y(\frac{dy}{dx})=0$

I don't know where to go from here, or even if the previous step is correct. Please help!

Edit: I have followed the advice given and I ended up with:

$\frac{x-2y}{2x-1}$

However this was incorrect. Someone please tell me what I am missing here.

${x}^{2}-4xy+{y}^{2}=4$

I know to take the derivatives of both sides, which would be:

$\frac{d}{dx}[{x}^{2}-4xy+{y}^{2}]=0$

I'm not sure if I did it right, but I then got:

$2x-4\cdot (x\frac{dy}{dx})+y+2y(\frac{dy}{dx})=0$

I don't know where to go from here, or even if the previous step is correct. Please help!

Edit: I have followed the advice given and I ended up with:

$\frac{x-2y}{2x-1}$

However this was incorrect. Someone please tell me what I am missing here.

asked 2022-08-11

I am trying to implicitly differentiate the following function:

$\lambda =\mathrm{exp}\left[{(\alpha +\frac{s}{\lambda -s})}^{-1}\right]$

Can someone help me with this?

$\lambda =\mathrm{exp}\left[{(\alpha +\frac{s}{\lambda -s})}^{-1}\right]$

Can someone help me with this?