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The problem is

$$y\,\sin\,x^3=x\,\sin\,y^3$$

Find the $y'$

The answer is enter image description here

Can some explain how to do this, please help.

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    Assume that $y$ is a function of $x$. That is $y = y(x)$. Thus $y' = y'(x)$ is just the derivative of $y$ (a function of $x$) with respect to $x$. Now you differentiate both sides of the equation with the rules you are used to. – 2017-01-19

4 Answers 4

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Differentiating with respect to $x$, we get :

$$y'\sin(x^3)+3x^2y\cos(x^3)=\sin(y^3)+3xy^2y'\cos(y^3)$$

At this point, we can easily compute $y'$ in terms of $x$ and $y$.

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Rewrite this as $$y(x)\sin{x^3}=x\sin{y(x)^3}$$ to see that everything depends on $x$. Now, take the derivative with respect to $x$ (as you know) and solve for $y'(x)$. In the end rewrite again $y'(x)$ as $y'$ and $y(x)$ as $y$ to conform with the usual notation:

\begin{align}0&=\frac{d}{dx}\left(y(x)\sin{x^3}-x\sin{y(x)^3}\right)\\[0.3cm]&=y'(x)\sin{x^3}+y(x)\cos{x^3}(x^3)'-\sin{y(x)^3}-x\cos{y(x)^3}(y(x)^3)'\\[0.2cm]&=y'(x)\sin{x^3}+3x^2y(x)\cos{x^3}-\sin{y(x)^3}-3xy(x)^2y'(x)\cos{y(x)^3}\\[0.3cm]\implies y'(x)&=\frac{\sin{y(x)^3}-3x^2y(x)\cos{x^3}}{\sin{x^3}-3xy(x)^2\cos{y(x)^3}}\\[0.3cm]\implies y'&=\frac{\sin{y^3}-3x^2y\cos{x^3}}{\sin{x^3}-3xy^2\cos{y^3}}\end{align} Now, you can repeat the same process by treating $x$ as a function of $y$. This will give you the second expression in the solution. But you do not need to do this.

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$y \sin x^3=x\sin y^3→$ Differentiating both sides with respect to $x .$

$⇒y(\cos x^3×3x^2)+y′\sin x^3 =\sin y^3+x \cos y^3×3y^2y′ $

$⇒3x^2y \cos x^3+y′\sin x^3=\sin y^3+3y^2x \cos y^3y′$

$⇒y′[\sin x^3−3y^2x\cos y^3]=\sin y^3−3x^2y \cos x^3$

$⇒y′=\dfrac{\sin y^3−3x^2y \cos x^3}{\sin x^3−3y^2x \cos y^3}$

I was searching it and I got the complete solution from here,

Implicit Differentiation

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I have found that it helps students to understand implicit differentiation if first they think of both $x$ and $y$ as functions of some third variable such as $t$ and take the derivative of both sides with respect to $t$, being careful to use the product rule, chain rule, etc when needed. Then as a final step, multiply both sides by $dt$. Since the object usually is to solve for $y^\prime=\frac{dy}{dx}$ you can then divide both sides by $dx$ and solve.

\begin{eqnarray} \frac{d}{dt}\left[y\sin(x^3)\right]&=&\frac{d}{dt}\left[x\sin(y^3)\right]\\ \sin(x^3)\frac{dy}{dt}+y\frac{d\sin(x^3)}{dt}&=&\sin(y^3)\frac{dx}{dt}+x\frac{d\sin(y^3)}{dt}\\ \sin(x^3)\frac{dy}{dt}+3x^2y\cos(x^3)\frac{dx}{dt}&=&\sin(y^3)\frac{dx}{dt}+3xy^2\cos(y^3)\frac{dy}{dt}\\ \sin(x^3)dy+3x^2y\cos(x^3)dx&=&\sin(y^3)dx+3xy^2\cos(y^3)dy\\ \sin(x^3)\frac{dy}{dx}+3x^2y\cos(x^3)&=&\sin(y^3)+3xy^2\cos(y^3)\frac{dy}{dx}\\ \left[\sin(x^3)-3xy^2\cos(y^3)\right]\frac{dy}{dx}&=&\sin(y^3)-3x^2y\cos(x^3)\\ \frac{dy}{dx}&=&\frac{\sin(y^3)-3x^2y\cos(x^3)}{\sin(x^3)-3xy^2\cos(y^3)} \end{eqnarray}

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    I'd like to add that it's not just pedagogically helpful, but it's also possible to use this approach to **rigorously** define Leibniz notation in terms of how the variables vary with respect to $t$ and with one another. Proofs still need some work to make rigorous, but the intuitive ideas like cancelling in the chain rule show up much more clearly in the proof. However, please you don't say "divide by $dx$" because such a thing is not rigorous. Rather, cite the chain rule. – 2017-01-19
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    @user21820 Yes, I hesitated to to say "divide by $dx$" but I have been spoiled by decades of working with engineers. :) But even when teaching future engineers I actually use $\Delta x$ at first. – 2017-01-19