1
$\begingroup$

I was studying Thomas's Calculus book and attempted a question using implicit differentiation.

$$x^3=\frac{2x-y}{x+3y}$$

I differentiated both sides directly, using the quotient rule on the RHS to obtain. $$ \frac{dy}{dx}=\frac{7y-3x^2(x+3y)^2}{7x} \hspace{1cm}(1)$$

The steps in the solution were to bring $x+3y$ to the LHS and differentiate every term to get $$ \frac{dy}{dx}=\frac{2-4x^3-9x^2y}{3x^3+1}\hspace{1cm} (2)$$

I was confused at first, and guessed that the two must be equal. And could show it using these steps. Expanding the numerator of $(1)$ gives $$\frac{dy}{dx}=\frac{7y-3x^4-18x^3y-27x^2y^2}{7x}\hspace{1cm} (1)$$

Replacing $x^3$ in the denominator of $(2)$ by $\frac{2x-y}{x+3y}$ and some simplification gives$$ \begin{align}\frac{dy}{dx}&=\frac{2x+6y-4x^4-21x^3y-27x^2y^2}{7x}\\&=\frac{7y-3x^4-18x^3-27x^2y^2+(2x-y-x^4-3x^3y)}{7x}\end{align}\hspace{1cm} (2)$$

The sum in the brackets is zero by $$2x-y-x^4-3x^3y=2x-y-x^3(x+3y)=2x-y-(2x-y)=0$$

Therefore $(1)=(2)$. Firstly, I am still not entirely sure $\textbf{why}$ we get two expressions which look completely different in the first place. Moreover is there any $\textbf{better way}$ to show that they are equal besides the fortuitous, trial and error, method I used? Many many thanks.

$\endgroup$
0
$\begingroup$

Solving your equation for $y$ we get $$y=\frac{2x-x^4}{3x^3+1}$$ now you can use the quotient rule.

$\endgroup$
  • $\begingroup$ Thanks, I did not notice that, I was wondering about the case when we can not solve for y. $\endgroup$ – matt Jan 2 at 12:24
  • $\begingroup$ You will need the condition $$3x^3+1\neq 0$$ $\endgroup$ – Dr. Sonnhard Graubner Jan 2 at 12:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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