Find the derivative of the following via implicit differentiation: $\frac{d}{dx}\left(y\right)=\frac{d}{dx}(\mathrm{cot}\left(5x\right))$

Using the chain rule, $\frac{d}{dx}\left(y\right)=\frac{dy\left(u\right)}{du}\frac{du}{dx}$, where $u=x$ and $\frac{d}{du}\left(y\left(u\right)\right)={y}^{\prime}\left(u\right):$ $\left(\frac{d}{dx}\left(x\right)\right){y}^{\prime}\left(x\right)=\frac{d}{dx}(\mathrm{cot}\left(5x\right))$

The derivative of $x$ is $1$: $1{y}^{\prime}\left(x\right)=\frac{d}{dx}(\mathrm{cot}\left(5x\right))$ INTERMEDIATE STEPS: Possible derivation: $\frac{d}{dx}\left(x\right)$ The limit definition of the derivative for a function $f(x)$ is $\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$. Let $f(x)=x$. Then: $f(x+h)=(x+h)$ Substitute $f(x+h)$ and $f(x)$ into the limit definition for the derivative: $\frac{d}{dx}\left(x\right)=\underset{h\to 0}{lim}\frac{(x+h)-x}{h}$ Combine like terms in the numerator: $=\underset{h\to 0}{lim}\frac{h}{h}$ The limit variable h approaches, but is never equal to, $0$ and therefore can be algebraically canceled: $=\underset{h\to 0}{lim}1$ The limit of any constant is that constant, so simplify to finish computing the derivative: $\frac{d}{dx}\left(x\right)=1$

Using the chain rule, $\frac{d}{dx}(\mathrm{cot}\left(5x\right))=\frac{d\mathrm{cot}\left(u\right)}{du}\frac{du}{dx}$, where $u=5x$ and $\frac{d}{du}(\mathrm{cot}\left(u\right))=-{\mathrm{csc}}^{2}\left(u\right)$: ${y}^{\prime}\left(x\right)=-{\mathrm{csc}}^{2}\left(5x\right)\left(\frac{d}{dx}\left(5x\right)\right)$ INTERMEDIATE STEPS: Possible derivation: