The derivative of $f$ at $x=c$ is the limit of the average rate of change as the interval shrinks to a point:
provided this limit exists. Geometrically $f'(c)$ is the slope of the tangent to the curve at $\big(c,f(c)\big)$. The function is differentiable at $c$ when this two-sided limit exists, which means the left-hand derivative and right-hand derivative are equal:
Differentiability implies continuity
If $f$ is differentiable at $c$, then it is continuous at $c$. The converse is false: $f(x)=|x|$ is continuous at $0$ but has a corner there, so $f'(0)$ does not exist (left slope $-1$, right slope $+1$). The practical lesson is that corners, kinks, vertical tangents and discontinuities all break differentiability — so a quick continuity check is the first filter, but passing it is not enough.
Standard derivatives
| $f(x)$ | $f'(x)$ | $f(x)$ | $f'(x)$ |
|---|---|---|---|
| $x^{n}$ | $nx^{n-1}$ | $\sin x$ | $\cos x$ |
| $\cos x$ | $-\sin x$ | $\tan x$ | $\sec^2 x$ |
| $\cot x$ | $-\csc^2 x$ | $\sec x$ | $\sec x\tan x$ |
| $\csc x$ | $-\csc x\cot x$ | $e^{x}$ | $e^{x}$ |
Exponential and logarithmic functions
Beyond $e^{x}$ and $\log x$, you need the general-base versions, all provable from the chain rule and the limit definition:
Here $\log$ denotes the natural logarithm $\ln$ (base $e$), as in NCERT. The factor $\log a$ appears whenever the base is not $e$ — it is the price of changing base.
Derivatives of inverse trigonometric functions
These follow by differentiating the defining relation (e.g. for $y=\sin^{-1}x$, $\sin y=x$) implicitly:
| $f(x)$ | $f'(x)$ | Valid for |
|---|---|---|
| $\sin^{-1}x$ | $\dfrac{1}{\sqrt{1-x^2}}$ | $-1 |
| $\cos^{-1}x$ | $-\dfrac{1}{\sqrt{1-x^2}}$ | $-1 |
| $\tan^{-1}x$ | $\dfrac{1}{1+x^2}$ | all $x$ |
| $\cot^{-1}x$ | $-\dfrac{1}{1+x^2}$ | all $x$ |
| $\sec^{-1}x$ | $\dfrac{1}{|x|\sqrt{x^2-1}}$ | $|x|>1$ |
| $\csc^{-1}x$ | $-\dfrac{1}{|x|\sqrt{x^2-1}}$ | $|x|>1$ |
Notice the pairing: each "co-" function's derivative is the negative of its partner's. A clever substitution (such as $x=\tan\theta$) often simplifies a messy inverse-trig expression to a single inverse function before you differentiate — this is the heart of the harder Ex 5.3 problems.
The chain rule
For a composite $y=f\big(g(x)\big)$,
Differentiate the outer function (leaving the inside alone), then multiply by the derivative of the inside. Combined with the product rule $(uv)'=u'v+uv'$ and the quotient rule $\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^2}$, the chain rule handles almost every function in this chapter, often peeling off one layer at a time.
Second-order derivatives
Differentiating $\dfrac{dy}{dx}$ again gives the second derivative $\dfrac{d^2y}{dx^2}$ (also written $y''$ or $f''(x)$) — the rate of change of the slope, which governs concavity. A common exam task is to prove a differential-equation relation by successive differentiation: differentiate the given $y$ once and twice, then substitute to verify an identity such as $y''+y=0$. The technique is to keep re-using the first relation to eliminate awkward terms.