Find $\frac{dy}{dx}$ given the equation $x^2y-e^y+9=0$

111 Views Asked by user637978 At 27 Mar 2026 - 9:59

Solution:

$\frac{d}{dx}(x^2y-e^y+9)=0$

$\frac{d}{dx}(x^2y)-\frac{d}{dx}(e^y)+\frac{d}{dx}9=0$

Now, just be to as confusing as possible, lets switch our notation from $\frac{d}{dx}$ to using $'$ notation. (For example, $\frac{d}{dx}x^2=(x^2)'$)

Keeping this transition of notation in mind and using the product rule on $x^2y$ and our derivative identity on $e^y$, we get:

$(x^2)'y+(x^2)y'-(e^y)y'=0$

Where $y'=\frac{dy}{dx}$. So we just need to solve this for $y'$

$2xy+(x^2)y'-(e^y)y'=0$

$(x^2-e^y)y'=-2xy$

$y'=\frac{-2xy}{x^2-e^y}$