I'm currently learning implicit differentiation (which I am having a lot of difficulties with) and I have encountered the following equation
I'm not exactly sure how we got from the third step to the fourth step. Could someone please explain it to me? Thanks in advance for any help!
Best Answer
Following my comment, given two functions $f, g$ of $x$, recall that the product rule is $$ \frac{\mathrm{d}}{\mathrm{d}x}\Big[\, f(x)g(x) \,\Big] = f'(x)g(x) + f(x)g'(x). $$ If $y$ is a function of $x$ and $y$ is unknown, we can just write $$ \frac{\mathrm{d}}{\mathrm{d}x}\Big[\, y \,\Big] = \frac{\mathrm{d}y}{\mathrm{d}x}. $$ Now, consider the derivative of $x^{3}y$. The first function is $x^{3}$ and the second function is $y$, so that $$ \frac{\mathrm{d}}{\mathrm{d}x}\Big[\, x^{3}y \,\Big] = \frac{\mathrm{d}}{\mathrm{d}x}\Big[\, x^{3} \,\Big]\cdot y + x^{3} \cdot \frac{\mathrm{d}}{\mathrm{d}x}\Big[\, y \,\Big] = 3x^{2}y + x^{3}\frac{\mathrm{d}y}{\mathrm{d}x}. $$
Also recall that the chain rule for $f$ and $g$ is $$ \frac{\mathrm{d}}{\mathrm{d}x}\Big[\, f\big(g(x)\big) \,\Big] = f'\big(g(x)\big) \cdot g'(x). $$ Consider the derivative of $y^{2}$. This time we use the chain rule with $f(y) = y^{2}$ and $g(x) = y$, so that $$ \frac{\mathrm{d}}{\mathrm{d}x}\Big[\, y^{2} \,\Big] = 2y \cdot \frac{\mathrm{d}}{\mathrm{d}x}\Big[\, y \,\Big] = 2y\frac{\mathrm{d}y}{\mathrm{d}x}. $$