2022-01-10

first derivative of the given function

$y=cot5x$

nick1337

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

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

The derivative of $x$ is $1$:
$1{y}^{\prime }\left(x\right)=\frac{d}{dx}\left(\mathrm{cot}\left(5x\right)\right)$
INTERMEDIATE STEPS:
Possible derivation:
$\frac{d}{dx}\left(x\right)$
The limit definition of the derivative for a function $f\left(x\right)$ is $\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{lim}\frac{f\left(x+h\right)-f\left(x\right)}{h}$. Let $f\left(x\right)=x$. Then:
$f\left(x+h\right)=\left(x+h\right)$
Substitute $f\left(x+h\right)$ and $f\left(x\right)$ into the limit definition for the derivative:
$\frac{d}{dx}\left(x\right)=\underset{h\to 0}{lim}\frac{\left(x+h\right)-x}{h}$
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, $\frac{d}{dx}\left(\mathrm{cot}\left(5x\right)\right)=\frac{d\mathrm{cot}\left(u\right)}{du}\frac{du}{dx}$, where $u=5x$ and $\frac{d}{du}\left(\mathrm{cot}\left(u\right)\right)=-{\mathrm{csc}}^{2}\left(u\right)$:
${y}^{\prime }\left(x\right)=-{\mathrm{csc}}^{2}\left(5x\right)\left(\frac{d}{dx}\left(5x\right)\right)$
INTERMEDIATE STEPS:
Possible derivation: