Foundations, the Chain Rule, and Differentiability

The standard derivatives you must recall instantly:

• $\dfrac{d}{dx}x^n = nx^{n-1}$, $\dfrac{d}{dx}\sqrt{x} = \dfrac{1}{2\sqrt{x}}$, $\dfrac{d}{dx}\dfrac{1}{x} = -\dfrac{1}{x^2}$
• $\dfrac{d}{dx}e^x = e^x$, and $\dfrac{d}{dx}a^x = a^x \log a$
• $\dfrac{d}{dx}\log x = \dfrac{1}{x}$, and $\dfrac{d}{dx}\log_a x = \dfrac{1}{x \log a}$
• $\dfrac{d}{dx}\sin x = \cos x$, $\dfrac{d}{dx}\cos x = -\sin x$, $\dfrac{d}{dx}\tan x = \sec^2 x$
• $\dfrac{d}{dx}\sec x = \sec x \tan x$, $\dfrac{d}{dx}\csc x = -\csc x \cot x$, $\dfrac{d}{dx}\cot x = -\csc^2 x$

The three combining rules:

• Sum/difference: $\dfrac{d}{dx}[u \pm v] = u' \pm v'$
• Product: $\dfrac{d}{dx}[u\,v] = u'v + uv'$
• Quotient: $\dfrac{d}{dx}\!\left(\dfrac{u}{v}\right) = \dfrac{u'v - uv'}{v^2}$

If $y = f(g(x))$, then $\dfrac{dy}{dx} = f'(g(x)) \cdot g'(x)$. For multiple nested layers, multiply the derivative of every layer:

• $\dfrac{d}{dx}[g(x)]^n = n[g(x)]^{n-1}\,g'(x)$
• $\dfrac{d}{dx}\sin[g(x)] = \cos[g(x)]\,g'(x)$, $\dfrac{d}{dx}\cos[g(x)] = -\sin[g(x)]\,g'(x)$
• $\dfrac{d}{dx}\log[g(x)] = \dfrac{g'(x)}{g(x)}$, $\dfrac{d}{dx}e^{g(x)} = e^{g(x)}\,g'(x)$

By the chain rule, $\dfrac{d}{dx}f(f(x)) = f'(f(x)) \cdot f'(x)$, and for three layers $\dfrac{d}{dx}f(f(f(x))) = f'(f(f(x))) \cdot f'(f(x)) \cdot f'(x)$. If a fixed point is given (e.g. $f(1) = 1$), each nested $f$ at that point is still $1$, so every factor becomes $f'(1)$. When an inner expression carries its own coefficient (such as $f(2f(x) + 2)$), the chain rule pulls out that extra factor too — do not drop it.

Before differentiating, look for cheap algebraic simplifications:

• Common factors in a quotient that cancel.
• Negative/fractional powers that combine — e.g. multiply top and bottom by the lower power to clear them.
• Identities that reduce a product or ratio to a standard form.

Only after the expression is in its simplest form do you apply the rules. This converts an ugly derivative into a routine one and removes most of the error surface.

For a small change $h$ about a point $a$: $f(a + h) \approx f(a) + h\,f'(a)$. Choose $a$ so that $f(a)$ is easy to compute exactly; let $h$ be the small (possibly negative) gap to the target. The term $h\,f'(a)$ is the tangent-line correction. The closer $h$ is to zero, the better the estimate.

The slope of the tangent to $y = f(x)$ at $x = a$ is $f'(a)$. To find where the slope itself is greatest or least, treat the slope function $s(x) = f'(x)$ as a new function and analyse IT: set $s'(x) = f''(x) = 0$ to locate the candidate points, then compare $s$-values. Equation of the tangent at $(a, f(a))$: $y - f(a) = f'(a)(x - a)$.

$f$ is differentiable at $x = a$ if the left-hand derivative equals the right-hand derivative:

• Differentiable $\Rightarrow$ continuous (but NOT the converse — $|x|$ is continuous yet not differentiable at $0$).
• A derivative typically fails at corners ($|x|$ at $0$), cusps, breaks (jump discontinuities), and vertical tangents.
• A modulus inside a product can be smoothed: if another factor vanishes at the corner, the product may be differentiable everywhere.

Keep these collapses in reflex memory:

• Half-angle of $1\pm\cos x$: $1-\cos x = 2\sin^2\tfrac{x}{2}$, $1+\cos x = 2\cos^2\tfrac{x}{2}$; so $\dfrac{1-\cos x}{1+\cos x}=\tan^2\tfrac{x}{2}$, $\dfrac{1}{1-\cos x}=\tfrac12\csc^2\tfrac{x}{2}$, $\dfrac{1}{1+\cos x}=\tfrac12\sec^2\tfrac{x}{2}$.
• Power-reduction (double angle): $1-\cos 2x = 2\sin^2 x$, $1+\cos 2x = 2\cos^2 x$, $\sin 2x = 2\sin x\cos x$.
• Perfect square under a root: $1\pm\sin 2x=(\sin x\pm\cos x)^2$ and $1\pm\sin\theta=\left(\cos\tfrac\theta2\pm\sin\tfrac\theta2\right)^2$, so $\sqrt{1\pm\sin 2x}=|\sin x\pm\cos x|$ — keep the modulus; its sign depends on the interval.
• $\sec\pm\tan$: $\sec x+\tan x=\dfrac{1+\sin x}{\cos x}=\tan\!\left(\tfrac\pi4+\tfrac x2\right)$, $\sec x-\tan x=\tan\!\left(\tfrac\pi4-\tfrac x2\right)$.
• Harmonic form: $a\sin x+b\cos x=\sqrt{a^2+b^2}\,\sin(x+\alpha)$, so its extreme values are $\pm\sqrt{a^2+b^2}$.
• Weierstrass $t=\tan\tfrac{x}{2}$: $\sin x=\dfrac{2t}{1+t^2}$, $\cos x=\dfrac{1-t^2}{1+t^2}$ — useful whenever a rational function of $\sin x,\cos x$ must be handled in one variable.