Find dy/dx 16e^y=x^4+y^5

To apply the Chain Rule, set ${\displaystyle u}$ as ${\displaystyle y}$.

Differentiate using the Exponential Rule which states that ${\displaystyle \frac{d}{du}\left[a^u\right]}$ is ${\displaystyle a^u\ln \left(a\right)}$ where ${\displaystyle a}$=${\displaystyle e}$.

Replace all occurrences of ${\displaystyle u}$ with ${\displaystyle y}$.

By the Sum Rule, the derivative of ${\displaystyle x^4+y^5}$ with respect to ${\displaystyle x}$ is ${\displaystyle \frac{d}{dx}\left[x^4\right]+\frac{d}{dx}\left[y^5\right]}$.

Differentiate using the Power Rule which states that ${\displaystyle \frac{d}{dx}\left[x^n\right]}$ is ${\displaystyle n{x}^{n-1}}$ where ${\displaystyle n=4}$.

Differentiate using the chain rule, which states that ${\displaystyle \frac{d}{dx}\left[f\left(g\left(x\right)\right)\right]}$ is ${\displaystyle f\prime \left(g\left(x\right)\right)g\prime \left(x\right)}$ where ${\displaystyle f\left(x\right)=x^5}$ and ${\displaystyle g\left(x\right)=y}$.

Rewrite ${\displaystyle \frac{d}{dx}\left[y\right]}$ as ${\displaystyle y\mathrm{<mo>\prime </mo>}}$.

Factor ${\displaystyle y\mathrm{<mo>\prime </mo>}}$ out of ${\displaystyle 16y\mathrm{<mo>\prime </mo>}{e}^y}$.

Factor ${\displaystyle y\mathrm{<mo>\prime </mo>}}$ out of ${\displaystyle -5y^{4}y\mathrm{<mo>\prime </mo>}}$.

Factor ${\displaystyle y\mathrm{<mo>\prime </mo>}}$ out of ${\displaystyle y\mathrm{<mo>\prime </mo>}\left(16e^y\right)+y\mathrm{<mo>\prime </mo>}(-5y^4)}$.

Divide each term in ${\displaystyle y\mathrm{<mo>\prime </mo>}(16e^y-5y^4)=4x^3}$ by ${\displaystyle 16e^y-5y^4}$.

Simplify the left side.