Q.Find the derivative of the following function with respect to the corresponding independent variable: $y = \dfrac{\sin x}{1+\cos x}$

Applying the first-principle limit once to each basic elementary function builds a permanent table; every later problem then differentiates by combining this table with the rules of differentiation (sum/product/quotient/chain/constant-multiple), with no further limit ever required.

Algebraic functions.

Logarithmic and exponential functions.

($e^x$ is the unique elementary function that is its own derivative — the special case $a=e$, $\log e=1$.)

The six trigonometric functions.

Only $\sin x\to\cos x$ is derived directly from the limit definition (via the sum-to-product identity and $\lim_{\theta\to0}\sin\theta/\theta=1$); every other trig derivative follows from it using the chain rule ($\cos x=\sin(\tfrac\pi2+x)$) or the quotient rule ($\tan x=\sin x/\cos x$, etc.) — so the whole trig table rests on a single limit.

The six inverse trigonometric functions (each domain-restricted to its principal branch):

$\frac{d}{dx}\!\left(\sin^{-1}x\right)=\frac{1}{\sqrt{1-x^2}},\qquad \frac{d}{dx}\!\left(\cos^{-1}x\right)=-\frac{1}{\sqrt{1-x^2}},\qquad \frac{d}{dx}\!\left(\tan^{-1}x\right)=\frac{1}{1+x^2},$ …