Calculus implicit differentiation question

I stumbled upon this calculus implicit differential question: find $$\cfrac{du}{dy}$$ of the function $$u = \sin(y^2+u)$$.

The answer is $$\cfrac{2y\cos(y^2+u)}{1−\cos(y^2+u)}$$. I understand how to get the answer for the numerator, but how do we get the denominator part?

And can anyone point out the real intuitive difference between chain rule and implicit differentiation? I can't seem to get my head around them and when or where should I use them.

We have $$u=\sin(y^2+u)$$, differentiating gives $$\begin{eqnarray*} \frac{du}{dy} =\left( 2y+ \frac{du}{dy} \right) \cos(y^2+u) \end{eqnarray*}$$ This can be rearranged to $$\begin{eqnarray*} \frac{du}{dy}(1-\cos(y^2+u)) =2y \cos(y^2+u) . \end{eqnarray*}$$

So $$\begin{eqnarray*} \frac{du}{dy} =\frac{ 2y \cos(y^2+u) }{1- \cos(y^2+u) }. \end{eqnarray*}$$

• But how do you get 1-cos(y^2+u)? Commented Dec 10, 2019 at 2:37
• Collect all the $du/dy$'s onto the LHS. I have put an extra line in to explain. Commented Dec 10, 2019 at 2:42
• Ah, I see. I guess I overworked things a bit. Thanks Donald! Commented Dec 10, 2019 at 2:48

When applying the chain rule, you need the differentiate $$u$$ also. Explicitly, $$\frac{\mathrm{d}u}{\mathrm{d}y} = \frac{\mathrm{d}}{\mathrm{d}y} \Big( \sin(y^2 + u) \Big) = \cos(y^2 + u) \Big( \frac{\mathrm{d}}{\mathrm{d}y} (y^2 + u) \Big) = \cos(y^2 + u) \Big( 2y + \frac{\mathrm{d}u}{\mathrm{d}y} \Big).$$ Solving for $$\frac{\mathrm{d}u}{\mathrm{d}y}$$, we find the answer you announced.

$$u = \sin(y^2+u) \implies\frac{du}{dy} = \cos(y^2+u)\times\left(2y+\frac{du}{dy}\right)$$ $$\implies \frac{du}{dy} (1-\cos(y^2+u)) = 2y\cos(y^2+u) \implies \frac{du}{dy}=\frac{2y\cos(y^2+u)}{1-\cos(y^2+u)}$$

Starting with:

$$u = \sin(y^2+u)$$

Perhaps more apparent and facile path (especially on an exam where an equivalent valid solution may suffice) would be just take the inverse of the sine function for both sides. This results here in a rapid and easy separation of $$u$$ and $$y$$:

$$\arcsin(u) = y^2+u$$

$$\arcsin(u) - u = y^2$$

Taking the differentials of both sides and applying the formula for differentiating the arc sine function:

$$du/\sqrt(1-u^2) - du = 2 y dy$$

Whereupon it is evident:

$$\cfrac{du}{dy} = \cfrac{2y}{1/\sqrt(1-u^2)-1}$$

which I claim as a likely equivalent and facile solution path. Note: the domain of the inverse sin function can be likely defined in a suitable fashion to accommodate real world conditions (see comments here).

I would like to discuss the relationship between implicit differentiation and the chain rule. Consider the following example:

$$x^2 + y^2 = 25$$

If we differentiate both sides, we are left with the following result:

$$\frac{\mathrm{d}}{\mathrm{d}x} \left ( x^2 \right ) + \frac{\mathrm{d}}{\mathrm{d}x} \left ( y^2 \right ) = \frac{\mathrm{d}}{\mathrm{d}x} \left ( 25 \right ) \implies 2x + \frac{\mathrm{d}}{\mathrm{d}x} \left ( y^2 \right ) = 0$$

The question, then, is what to do with $$\frac{\mathrm{d}}{\mathrm{d}x} \left ( y^2 \right )$$ since, in general, there isn't an easy way to express the relationship between $$x$$ and $$y$$. To tackle this problem, we'll pretend that there is a relationship. Explicitly, assume $$y = f(x)$$ for some unknown function.

We reframe the problem as follows:

$$\frac{\mathrm{d}}{\mathrm{d}x} \left ( y^2 \right ) = \frac{\mathrm{d}}{\mathrm{d}x} \left ( f^2(x) \right ) = 2 f(x) \times \frac{\mathrm{d}f}{\mathrm{d}x} = 2 y \left ( \frac{\mathrm{d}y}{\mathrm{d}x} \right )$$

Hope this helped!