Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Assume that $x$ and $y$ are related by the equation $y\ln x=e^{1−x}+y^3$. Compute $dy/dx$ evaluated at $x=1$.

I do not understand how to compute the derivative of an equation. Please explain.

share|cite|improve this question
Are you familiar with implicit differentiation? –  Kaj Hansen Jun 23 '14 at 2:17
You can't find the derivative of an equation, but you can find the derivative of each side of the equation. And if the two sides were equal, then their derivatives will be equal. –  David Jun 23 '14 at 2:39
It should be noted, the reason implicit differentiation is "ok" is that the Implicit Function Theorem states that given a certain condition we can solve for one variable as a function of the remaining variable. Without that condition, it may or may not be possible to solve for $y$ as a function of $x$ and thus it may not make sense to differentiate w.r.t. $x$ at such a point. However, you can always take the total differential of an equation and that has meaning... but, this is a calculus I question so I leave this as a comment for your future. –  James S. Cook Jun 23 '14 at 2:50

6 Answers 6

up vote 1 down vote accepted

As the others have mentioned, your problem requires implicit differentiation. Firstly, anytime you differentiate, you are secret implicitly deriving but you just don't know it! Your problem is the following:

$$y\ln x=e^{1−x}+y^3$$

and we are required to find $\frac{dy}{dx}$ which means find the derivative with respect to $x$. This means, $y$ is a function $x$ or in other words, $y(x)$. So, let's get to differentiating.

\begin{align} y\ln x &= e^{1-x} + y^3 \\ \frac{d}{dx} y\ln x &= \frac{d}{dx} e^{1-x} + \frac{d}{dx} y^3 \\ \end{align}

Now, because $y$ is actually a function it's derivative should be $\frac{dy}{dx}$. Keeping that in mind, let's continue. On the left hand side, we'll use the product rule and on the right hand side, we'll differentiate seprately using chain rule and product rule. Finally, we'll rearrange for $\frac{dy}{dx}$.

\begin{align} \frac{d}{dx} y\ln x &= \frac{d}{dx} e^{1-x} + \frac{d}{dx} y^3 \\ \frac{dy}{dx}\ln x + y\left(\frac{1}{x}\right) &= e^{1-x} \cdot (-1) + \frac{dy}{dx}\cdot3y^2 \\ \frac{dy}{dx}\ln x - \frac{dy}{dx}3y^2 &= -\left(e^{1-x} + \frac{y}{x}\right)\\ \end{align}

I'm going to leave the last few steps for you to do but it's almost done! All you have to do is isolate $\frac{dy}{dx}$ and you're done!

If you have any questions then please don't hesitate to ask!

Written with StackEdit.

share|cite|improve this answer
@downvoter, may I know the reason so I can improve the post? :) thanks! –  Jeel Shah Jul 5 '14 at 4:02

Here are the first few steps you need to do. As others have noted, it is called implicit differentiation. It is very similar to "normal" differentiation, but the difference is that you consider $y$ to be a function of $x$. When you want to differentiate it with respect to $x$, you get that $\frac{d}{dx}(y) = \frac{dy}{dx}$ (very simply) . Also, you need to remember to apply to $y$ all of the rules you normally apply to functions of $x$. For example, you need to apply the product rule to $\frac{d}{dx}(xy)$, among other things.

So you have that

$$y\ln x = e^{1-x} + y^3$$ thus $$\frac{d}{dx} (y\ln x) = \frac{d}{dx} ( e^{1-x} + y^3 )$$ so $$\frac{dy}{dx}\ln x + \frac{y}{x} = -e^{1-x} + 3y^2 \frac{dy}{dx}$$

Now you should be able to solve for $\frac{dy}{dx}$, which will be dependent of both $x$ and $y$.

share|cite|improve this answer
Good answer, +1, but please: "differentiate", "differentiation", not derivate and derivation. –  David Jun 23 '14 at 2:42
Thanks for the comment, my english sometimes needs improvements. –  Olivier Jun 23 '14 at 2:43

Use the chain rule/implicit differentiation. For example, if $f(x)s(y)+g(x)+h(y)=j(y)$ is an equation ($y=q(x)$ a function of $x$), then differentiate both sides to obtain $$f^\prime(x)s(y)+f(x)s^\prime(y)y^\prime+g^\prime(x)+h^\prime(y)y^\prime=j^\prime(y)y^\prime\\ \implies f^\prime(x)s(y)+g^\prime(x)=y^\prime(j^\prime(y)-h^\prime(y)-f(x)f^\prime(y))\\ \implies y^\prime=\dfrac{f(x)s^\prime(y)+g^\prime(x)}{j^\prime(y)-h^\prime(y)-f(x)s^\prime(y)}$$ In your case, $f(x)=\ln x,s(y)=y, h(y)=-y^3, g(x)=-e^{1-x},$ and $j(y)=0$.

share|cite|improve this answer

You need to do it implicitly. You noted that you don't know how to differentiate an equation, but what you want to do is find the quantity dy/dx. To do this, note that, if you differentiate both expressions, they will remain equal

share|cite|improve this answer

Hint: You're not computing the derivative of an equation. You're computing the derivative of $y$ with respect to $x$. Since you aren't given $y$ in terms of $x$, your answer will most likely be in terms of both $y$ and $x$, not just $x$.

share|cite|improve this answer

Here is the complete answer. First differentiate both sides of the equation $$ D(y \ln(x)) = D(e^{1-x}+y^3)\\ \to D(\ln(x))\times y+ D(y)\times \ln(x) = -e^{1-x} + D(y^3) $$ Using implicit differentiation we have $$ \frac{1}{x}y+y^{\prime} \ln(x) = -e^{1-x}+3y^2 y^{\prime} $$ Now we have $$ y^{\prime} \ln(x) - 3y^2 y^{\prime} +\frac{1}{x}y + e^{1-x} = 0 \\ \to y^{\prime}(\ln(x)-3y^2)=-\frac{1}{x}y-e^{1-x} $$ Thus $$ y^{\prime}=\frac{-y/x-e^{1-x}}{\ln(x)-3y^2} $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.