Given ${e}^{L}+KL=K{e}^{K}$, we are being asked to find $\frac{dL}{dK}$. I think O need to use implicit differentiation, but I am really not sure how to do it!

Deanna Gregory
2022-09-28
Answered

Given ${e}^{L}+KL=K{e}^{K}$, we are being asked to find $\frac{dL}{dK}$. I think O need to use implicit differentiation, but I am really not sure how to do it!

You can still ask an expert for help

Lichtpulsax

Answered 2022-09-29
Author has **4** answers

Consider the implicit function

$f(K,L)={e}^{L}+KL-K{e}^{K}=0$

Now

$\frac{\mathrm{\partial}f(K,L)}{\mathrm{\partial}K}=L-{e}^{K}(K+1)\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}\frac{\mathrm{\partial}f(K,L)}{\mathrm{\partial}L}={e}^{L}+K$

Now, using the implicit function theorem

$\frac{dL}{dK}=-\frac{\frac{\mathrm{\partial}f(K,L)}{\mathrm{\partial}K}}{\frac{\mathrm{\partial}f(K,L)}{\mathrm{\partial}L}}=\frac{{e}^{K}(K+1)-L}{K+{e}^{L}}$

$f(K,L)={e}^{L}+KL-K{e}^{K}=0$

Now

$\frac{\mathrm{\partial}f(K,L)}{\mathrm{\partial}K}=L-{e}^{K}(K+1)\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}\frac{\mathrm{\partial}f(K,L)}{\mathrm{\partial}L}={e}^{L}+K$

Now, using the implicit function theorem

$\frac{dL}{dK}=-\frac{\frac{\mathrm{\partial}f(K,L)}{\mathrm{\partial}K}}{\frac{\mathrm{\partial}f(K,L)}{\mathrm{\partial}L}}=\frac{{e}^{K}(K+1)-L}{K+{e}^{L}}$

emmostatwf

Answered 2022-09-30
Author has **2** answers

There are a lot of different ways to do implicit differentiation. Some methods have you differentiate respect to a particular variable, but, the way I prefer is to just do differentiation without being with respect to a particular variable, and then solve for the derivative you are looking for. So, instead of the derivative rule for $y={u}^{n}$ being $\frac{dy}{du}=n{u}^{n-1}$, I instead have the differential rule $dy=n{u}^{n-1}\phantom{\rule{thinmathspace}{0ex}}du$. Then I can algebraically divide by $du$ if I want to find $\frac{dy}{du}$.

So, in your case, we have:

${e}^{L}+KL=K{e}^{K}$

Differentiating both sides gives us:

$d({e}^{L}+KL)=d\left(K{e}^{K}\right)\phantom{\rule{0ex}{0ex}}d\left({e}^{L}\right)+d\left(KL\right)=d\left(K{e}^{K}\right)\phantom{\rule{0ex}{0ex}}{e}^{L}\phantom{\rule{thinmathspace}{0ex}}dL+K\phantom{\rule{thinmathspace}{0ex}}dL+L\phantom{\rule{thinmathspace}{0ex}}dK=K{e}^{K}\phantom{\rule{thinmathspace}{0ex}}dK+{e}^{K}\phantom{\rule{thinmathspace}{0ex}}dK$

now we just need to gather our terms together so we can solve for $\frac{dL}{dK}$:

${e}^{L}\phantom{\rule{thinmathspace}{0ex}}dL+K\phantom{\rule{thinmathspace}{0ex}}dL+L\phantom{\rule{thinmathspace}{0ex}}dK=K{e}^{K}\phantom{\rule{thinmathspace}{0ex}}dK+{e}^{K}\phantom{\rule{thinmathspace}{0ex}}dK\phantom{\rule{0ex}{0ex}}{e}^{L}\phantom{\rule{thinmathspace}{0ex}}dL+K\phantom{\rule{thinmathspace}{0ex}}dL=K{e}^{K}\phantom{\rule{thinmathspace}{0ex}}dK+{e}^{K}\phantom{\rule{thinmathspace}{0ex}}dK-L\phantom{\rule{thinmathspace}{0ex}}dK\phantom{\rule{0ex}{0ex}}({e}^{L}+K)\phantom{\rule{thinmathspace}{0ex}}dL=(K{e}^{K}+{e}^{K}-L)\phantom{\rule{thinmathspace}{0ex}}dK\phantom{\rule{0ex}{0ex}}\frac{dL}{dK}=\frac{K{e}^{K}+{e}^{K}-L}{{e}^{L}+K}$

So, in your case, we have:

${e}^{L}+KL=K{e}^{K}$

Differentiating both sides gives us:

$d({e}^{L}+KL)=d\left(K{e}^{K}\right)\phantom{\rule{0ex}{0ex}}d\left({e}^{L}\right)+d\left(KL\right)=d\left(K{e}^{K}\right)\phantom{\rule{0ex}{0ex}}{e}^{L}\phantom{\rule{thinmathspace}{0ex}}dL+K\phantom{\rule{thinmathspace}{0ex}}dL+L\phantom{\rule{thinmathspace}{0ex}}dK=K{e}^{K}\phantom{\rule{thinmathspace}{0ex}}dK+{e}^{K}\phantom{\rule{thinmathspace}{0ex}}dK$

now we just need to gather our terms together so we can solve for $\frac{dL}{dK}$:

${e}^{L}\phantom{\rule{thinmathspace}{0ex}}dL+K\phantom{\rule{thinmathspace}{0ex}}dL+L\phantom{\rule{thinmathspace}{0ex}}dK=K{e}^{K}\phantom{\rule{thinmathspace}{0ex}}dK+{e}^{K}\phantom{\rule{thinmathspace}{0ex}}dK\phantom{\rule{0ex}{0ex}}{e}^{L}\phantom{\rule{thinmathspace}{0ex}}dL+K\phantom{\rule{thinmathspace}{0ex}}dL=K{e}^{K}\phantom{\rule{thinmathspace}{0ex}}dK+{e}^{K}\phantom{\rule{thinmathspace}{0ex}}dK-L\phantom{\rule{thinmathspace}{0ex}}dK\phantom{\rule{0ex}{0ex}}({e}^{L}+K)\phantom{\rule{thinmathspace}{0ex}}dL=(K{e}^{K}+{e}^{K}-L)\phantom{\rule{thinmathspace}{0ex}}dK\phantom{\rule{0ex}{0ex}}\frac{dL}{dK}=\frac{K{e}^{K}+{e}^{K}-L}{{e}^{L}+K}$

asked 2022-08-16

So I got this classic implicit differentiation problem$y={x}^{x}$

I've tried to tackle it by transforming ${x}^{x}$ to the equivalent form of ${e}^{(\mathrm{ln}x)x}$:

$y={x}^{x}$

$y=({e}^{\mathrm{ln}x}{)}^{x}$

$y={e}^{(\mathrm{ln}x)x}$

and then proceeded to differentiate it in the following way:

${y}^{\prime}={e}^{(\mathrm{ln}x)x}\cdot \mathrm{ln}x\cdot \frac{1}{x}$

I've tried to tackle it by transforming ${x}^{x}$ to the equivalent form of ${e}^{(\mathrm{ln}x)x}$:

$y={x}^{x}$

$y=({e}^{\mathrm{ln}x}{)}^{x}$

$y={e}^{(\mathrm{ln}x)x}$

and then proceeded to differentiate it in the following way:

${y}^{\prime}={e}^{(\mathrm{ln}x)x}\cdot \mathrm{ln}x\cdot \frac{1}{x}$

asked 2022-11-17

I need to find the derivatives of

$(1+xy{)}^{3}=y{x}^{-1}$

I tried this and stuck

$3(1+xy{)}^{2}(0+x(\frac{dy}{dx})+y)=-y{x}^{-2}+({x}^{-1}\frac{dy}{dx})$

$\frac{dy}{dx}(x+y-{x}^{-1})=-y{x}^{-2}-3(1+xy{)}^{2}$

$\frac{dy}{dx}=\frac{-y{x}^{2}-3(1+xy{)}^{2}}{x+y-{x}^{-1}}$

I'm not quite sure how to simplify it to achieve the final answer -

$\frac{-y(3{x}^{2}(1+xy{)}^{2}+1}{x(3{x}^{2}(1+xy{)}^{2}-1}$

$(1+xy{)}^{3}=y{x}^{-1}$

I tried this and stuck

$3(1+xy{)}^{2}(0+x(\frac{dy}{dx})+y)=-y{x}^{-2}+({x}^{-1}\frac{dy}{dx})$

$\frac{dy}{dx}(x+y-{x}^{-1})=-y{x}^{-2}-3(1+xy{)}^{2}$

$\frac{dy}{dx}=\frac{-y{x}^{2}-3(1+xy{)}^{2}}{x+y-{x}^{-1}}$

I'm not quite sure how to simplify it to achieve the final answer -

$\frac{-y(3{x}^{2}(1+xy{)}^{2}+1}{x(3{x}^{2}(1+xy{)}^{2}-1}$

asked 2022-09-28

An equation is defined as $x+y+{x}^{5}-{y}^{5}=0$.

$a$. Show that the equation determines $y$ as a function of $x$ in a neighbourhood of the origin (0,0).

$b$. Denote the function from $(a)$ by $y=\phi (x)$. Find ${\phi}^{(5)}(0)$ and ${\phi}^{(2004)}(0)$.

For the first part,I simply needed to prove that the equation is ${C}^{1}$ in the neighbourhood of the origin and that the first derivative w.r.t.y does not vanish at that point. These are the sufficient conditions for the existence of an implicit function in the neighbourhood of the origin.

For the second part, I could do implicit differentiation (in a usual way) had it been for lower orders. But I need to learn the case for higher orders like the question asks. Please give me some hints.

$a$. Show that the equation determines $y$ as a function of $x$ in a neighbourhood of the origin (0,0).

$b$. Denote the function from $(a)$ by $y=\phi (x)$. Find ${\phi}^{(5)}(0)$ and ${\phi}^{(2004)}(0)$.

For the first part,I simply needed to prove that the equation is ${C}^{1}$ in the neighbourhood of the origin and that the first derivative w.r.t.y does not vanish at that point. These are the sufficient conditions for the existence of an implicit function in the neighbourhood of the origin.

For the second part, I could do implicit differentiation (in a usual way) had it been for lower orders. But I need to learn the case for higher orders like the question asks. Please give me some hints.

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?

asked 2022-07-22

use implicit differentiation to find $\frac{dy}{dx}$ in terms of $x$ and $y$

1.) ${x}^{3}-xy+{y}^{2}=4$

2.) $y=\mathrm{sin}(xy)$

find $\frac{{d}^{2}y}{d{x}^{2}}$

3.) ${x}^{2}{y}^{2}-2x=3$

1.) ${x}^{3}-xy+{y}^{2}=4$

2.) $y=\mathrm{sin}(xy)$

find $\frac{{d}^{2}y}{d{x}^{2}}$

3.) ${x}^{2}{y}^{2}-2x=3$

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How do you differentiate $x{e}^{{x}^{2}+{y}^{2}}$

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Use the process of implicit differentiation to find $\frac{dx}{dy}$

given that $\mathrm{sin}(x)+\mathrm{sin}(3y)=1$

given that $\mathrm{sin}(x)+\mathrm{sin}(3y)=1$