Block A — The Toolkit in Action

Example 1 — Roots and negative powers together

Differentiate $\sqrt{3x + 2} + \dfrac{1}{\sqrt{2x^2 + 4}}$ w.r.t. $x$.

Step 1 — rewrite as powers: $y = (3x+2)^{1/2} + (2x^2+4)^{-1/2}$.

Step 2 — chain rule on each: $$y' = \frac{1}{2}(3x+2)^{-1/2}(3) - \frac{1}{2}(2x^2+4)^{-3/2}(4x)$$

Answer: $\dfrac{3}{2\sqrt{3x+2}} - \dfrac{2x}{(2x^2+4)^{3/2}}$, defined for $x > -\frac{2}{3}$.

Example 2 — A change of base

Differentiate $\log_7(\log x)$, $x > 1$, w.r.t. $x$.

Step 1 — change base: $\log_7(\log x) = \dfrac{\log(\log x)}{\log 7}$ — the denominator is a constant.

Step 2 — differentiate the numerator: $\frac{d}{dx}\log(\log x) = \frac{1}{x\log x}$.

Answer: $\dfrac{1}{x\,\log 7\,\log x}$.

Example 3 — Simplify before touching the derivative

Differentiate $\cos^{-1}(\sin x)$ w.r.t. $x$.

Step 1 — co-function rewrite: $\sin x = \cos\left(\frac{\pi}{2} - x\right)$, so $f(x) = \cos^{-1}\cos\left(\frac{\pi}{2} - x\right) = \frac{\pi}{2} - x$ (on the principal range).

Step 2 — differentiate the linear result: $f'(x) = -1$.

Answer: $-1$ — a constant slope from an innocent-looking composite; brute chain rule gives $\frac{-\cos x}{\sqrt{1 - \sin^2 x}}$, which simplifies to the same.

Example 4 — The half-angle collapse inside an inverse tangent

Differentiate $\tan^{-1}\left(\dfrac{\sin x}{1 + \cos x}\right)$ w.r.t. $x$.

Step 1 — half-angle: $\dfrac{\sin x}{1 + \cos x} = \dfrac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}} = \tan\frac{x}{2}$.

Step 2 — collapse and differentiate: $f(x) = \tan^{-1}\tan\frac{x}{2} = \frac{x}{2}$, so $f'(x) = \frac{1}{2}$.

Answer: $\dfrac{1}{2}$.

Example 5 — An exponential hiding a double angle

Differentiate $\sin^{-1}\left(\dfrac{2^{x+1}}{1 + 4^x}\right)$ w.r.t. $x$ (for $x < 0$).

Step 1 — spot the pattern: $\dfrac{2^{x+1}}{1 + 4^x} = \dfrac{2 \cdot 2^x}{1 + (2^x)^2}$ — the double-angle shape with $2^x$ in place of $\tan\theta$.

Step 2 — collapse: $f(x) = 2\tan^{-1}(2^x)$ on the given range.

Step 3 — differentiate: $f'(x) = \dfrac{2 \cdot 2^x\log 2}{1 + 4^x} = \dfrac{2^{x+1}\log 2}{1 + 4^x}$.

Answer: as displayed — the substitution dictionary works even when the "$x$" inside is itself an exponential.

Example 6 — Sine to the power sine

Find $f'(x)$ if $f(x) = (\sin x)^{\sin x}$ for $0 < x < \pi$.

Step 1 — log: $\log y = \sin x \log(\sin x)$.

Step 2 — product + chain: $\frac{1}{y}y' = \cos x\log(\sin x) + \sin x \cdot \frac{\cos x}{\sin x} = \cos x\left[1 + \log(\sin x)\right]$.

Answer: $f'(x) = (\sin x)^{\sin x}\cos x\left[1 + \log(\sin x)\right]$.

Example 7 — Differentiating one function w.r.t. another

Differentiate $\sin^2 x$ with respect to $e^{\cos x}$.

Step 1 — the ratio rule: with $u = \sin^2 x$ and $v = e^{\cos x}$: $\frac{du}{dv} = \frac{du/dx}{dv/dx}$.

Step 2 — both rates: $\frac{du}{dx} = 2\sin x\cos x$ and $\frac{dv}{dx} = -\sin x\, e^{\cos x}$.

Step 3 — divide: $\frac{du}{dv} = \frac{2\sin x\cos x}{-\sin x\,e^{\cos x}} = -2\cos x\, e^{-\cos x}$.

Answer: $-2\cos x\,e^{-\cos x}$ — "differentiate $u$ w.r.t. $v$" always means divide the two $x$-derivatives.

Example 8 — A ninth power, no expansion

Differentiate $(3x^2 - 9x + 5)^9$ w.r.t. $x$.

Step 1 — chain rule: $9(3x^2 - 9x + 5)^8 \cdot (6x - 9)$.

Answer: $27(2x - 3)(3x^2 - 9x + 5)^8$ — factor the $3$ out of $6x - 9$ for the tidy final form.

Example 9 — Powers of sine and cosine

Differentiate $\sin^3 x + \cos^6 x$ w.r.t. $x$.

Step 1 — chain rule on each power: $3\sin^2 x\cos x + 6\cos^5 x(-\sin x)$.

Answer: $3\sin x\cos x\left(\sin x - 2\cos^4 x\right)$ — extract the common $3\sin x\cos x$ to finish cleanly.

Block B — Logarithmic and Inverse-Trig Classics

Example 10 — A trig exponent on a linear base

Differentiate $(5x)^{3\cos 2x}$ w.r.t. $x$.

Step 1 — log: $\log y = 3\cos 2x \log 5x$.

Step 2 — product rule: $\frac{1}{y}y' = -6\sin 2x\log 5x + \frac{3\cos 2x}{x}$.

Answer: $\dfrac{dy}{dx} = (5x)^{3\cos 2x}\left[\dfrac{3\cos 2x}{x} - 6\sin 2x\log 5x\right]$.

Example 11 — Three-halves power inside an inverse sine

Differentiate $\sin^{-1}(x\sqrt{x})$, $0 \leq x \leq 1$, w.r.t. $x$.

Step 1 — rewrite the inside: $x\sqrt{x} = x^{3/2}$.

Step 2 — chain rule: $\dfrac{1}{\sqrt{1 - x^3}} \cdot \dfrac{3}{2}x^{1/2}$.

Answer: $\dfrac{3\sqrt{x}}{2\sqrt{1 - x^3}}$ — note $(x^{3/2})^2 = x^3$ under the root.

Example 12 — A quotient with an inverse cosine upstairs

Differentiate $\dfrac{\cos^{-1}\frac{x}{2}}{\sqrt{2x + 7}}$, $-2 < x < 2$, w.r.t. $x$.

Step 1 — quotient rule pieces: $\frac{d}{dx}\cos^{-1}\frac{x}{2} = \frac{-1}{\sqrt{4 - x^2}}$ (chain rule brings a $\frac{1}{2}$ that merges into the root) and $\frac{d}{dx}\sqrt{2x+7} = \frac{1}{\sqrt{2x+7}}$.

Step 2 — assemble: $$\frac{dy}{dx} = \frac{\frac{-\sqrt{2x+7}}{\sqrt{4 - x^2}} - \frac{\cos^{-1}\frac{x}{2}}{\sqrt{2x+7}}}{2x + 7}$$

Answer: $-\left[\dfrac{1}{\sqrt{4 - x^2}\sqrt{2x+7}} + \dfrac{\cos^{-1}\frac{x}{2}}{(2x+7)^{3/2}}\right]$ — both terms negative, as expected for a decreasing numerator over a growing denominator.

Example 13 — The rationalising cotangent

Differentiate $\cot^{-1}\left[\dfrac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}\right]$, $0 < x < \dfrac{\pi}{2}$, w.r.t. $x$.

Step 1 — half-angle inside the roots: $1 \pm \sin x = \left(\cos\frac{x}{2} \pm \sin\frac{x}{2}\right)^2$, and on $\left(0, \frac{\pi}{2}\right)$ both brackets are positive, so the roots strip cleanly.

Step 2 — simplify the big fraction: $\dfrac{2\cos\frac{x}{2}}{2\sin\frac{x}{2}} = \cot\frac{x}{2}$.

Step 3 — collapse: $y = \cot^{-1}\cot\frac{x}{2} = \frac{x}{2}$, so $y' = \frac{1}{2}$.

Answer: $\dfrac{1}{2}$ — the entire beast was $\frac{x}{2}$ in disguise.

Example 14 — Log to the power log

Differentiate $(\log x)^{\log x}$, $x > 1$, w.r.t. $x$.

Step 1 — log: $\log y = \log x \cdot \log(\log x)$.

Step 2 — product rule: $\frac{1}{y}y' = \frac{1}{x}\log(\log x) + \log x \cdot \frac{1}{x\log x} = \frac{1}{x}\left[\log(\log x) + 1\right]$.

Answer: $\dfrac{dy}{dx} = (\log x)^{\log x}\cdot\dfrac{1 + \log(\log x)}{x}$.

Example 15 — Constants riding inside a cosine

Differentiate $\cos(a\cos x + b\sin x)$ w.r.t. $x$, for constants $a, b$.

Step 1 — chain rule: $-\sin(a\cos x + b\sin x) \cdot (-a\sin x + b\cos x)$.

Answer: $(a\sin x - b\cos x)\sin(a\cos x + b\sin x)$ — the two minus signs merge.

Example 16 — A sum that is secretly constant

Find $\dfrac{dy}{dx}$ if $y = \sin^{-1} x + \sin^{-1}\sqrt{1 - x^2}$, $0 < x < 1$.

Step 1 — recognise the pair: for $0 < x < 1$, $\sqrt{1 - x^2} = \cos(\sin^{-1}x)$, so the second term is $\cos^{-1} x$ — and $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$.

Step 2 — differentiate the constant: $0$. (Term-by-term: $\frac{1}{\sqrt{1-x^2}} + \frac{1}{\sqrt{1 - (1-x^2)}}\cdot\frac{-x}{\sqrt{1-x^2}} = \frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-x^2}} = 0$ ✓.)

Answer: $\dfrac{dy}{dx} = 0$.

Example 17 — An implicit relation that solves itself

If $x\sqrt{1 + y} + y\sqrt{1 + x} = 0$ for $-1 < x < 1$ (with $x \neq y$), prove that $\dfrac{dy}{dx} = -\dfrac{1}{(1 + x)^2}$.

Step 1 — isolate cleverly: $x\sqrt{1+y} = -y\sqrt{1+x}$; square both sides: $x^2(1 + y) = y^2(1 + x)$.

Step 2 — factor: $x^2 - y^2 + x^2 y - xy^2 = (x - y)(x + y + xy) = 0$; since $x \neq y$, we get $y = -\dfrac{x}{1 + x}$.

Step 3 — differentiate the explicit form (quotient rule): $$\frac{dy}{dx} = -\frac{(1 + x) - x}{(1 + x)^2} = -\frac{1}{(1 + x)^2}$$

Answer: proved. ∎ Squaring converted an awkward radical relation into a factorable polynomial — then the explicit function did the rest.

Block C — Mixed Classics and Deeper Results

Example 18 — Four terms, four different rules

Differentiate $x^x + x^a + a^x + a^a$ (for fixed $a > 0$, $x > 0$) w.r.t. $x$.

Step 1 — classify each term: variable-variable, variable-constant, constant-variable, constant-constant.

Step 2 — apply the right rule to each: $$\frac{dy}{dx} = x^x(1 + \log x) + a\,x^{a-1} + a^x\log a + 0$$

Answer: as displayed — one expression that examines the entire classification table at once.

Example 19 — Sine minus cosine to its own power

Differentiate $(\sin x - \cos x)^{(\sin x - \cos x)}$, $\dfrac{\pi}{4} < x < \dfrac{3\pi}{4}$, w.r.t. $x$.

Step 1 — the interval matters: on it, $\sin x - \cos x > 0$, so the logarithm is legal.

Step 2 — log and differentiate: $\log y = (\sin x - \cos x)\log(\sin x - \cos x)$; with $u = \sin x - \cos x$ and $u' = \cos x + \sin x$: $$\frac{1}{y}y' = u'\log u + u \cdot \frac{u'}{u} = (\cos x + \sin x)\left[1 + \log(\sin x - \cos x)\right]$$

Answer: $\dfrac{dy}{dx} = (\sin x - \cos x)^{(\sin x - \cos x)}(\cos x + \sin x)\left[1 + \log(\sin x - \cos x)\right]$.

Example 20 — A curvature-flavoured constant

If $(x - a)^2 + (y - b)^2 = c^2$ for some $c > 0$, prove that $\dfrac{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3/2}}{\frac{d^2y}{dx^2}}$ is a constant independent of $a$ and $b$.

Step 1 — first implicit differentiation: $(x - a) + (y - b)y' = 0$, so $y' = -\frac{x - a}{y - b}$.

Step 2 — second differentiation: $1 + (y')^2 + (y - b)y'' = 0$, so $y'' = -\frac{1 + (y')^2}{y - b}$.

Step 3 — assemble: $1 + (y')^2 = \frac{(x-a)^2 + (y-b)^2}{(y-b)^2} = \frac{c^2}{(y-b)^2}$; then $$\frac{\left[1 + (y')^2\right]^{3/2}}{y''} = \frac{c^3/(y-b)^3}{-c^2/(y-b)^3} = -c$$

Answer: the ratio equals $-c$, free of $a$ and $b$. ∎ (Geometrically: the expression measures the radius of curvature, and a circle's is its radius everywhere.)

Example 21 — An elegant implicit identity

If $\cos y = x\cos(a + y)$ with $\cos a \neq \pm 1$, prove that $\dfrac{dy}{dx} = \dfrac{\cos^2(a + y)}{\sin a}$.

Step 1 — differentiate implicitly: $-\sin y \, y' = \cos(a + y) - x\sin(a + y)\,y'$.

Step 2 — substitute $x = \frac{\cos y}{\cos(a+y)}$ and collect: $$y'\left[\frac{\cos y \sin(a+y)}{\cos(a+y)} - \sin y\right] = \cos(a+y)$$ The bracket is $\frac{\sin(a + y)\cos y - \cos(a + y)\sin y}{\cos(a+y)} = \frac{\sin a}{\cos(a+y)}$.

Step 3 — solve: $y' = \dfrac{\cos^2(a+y)}{\sin a}$. ∎

Answer: proved — the compound-angle formula $\sin(A - B)$ collapses the bracket in one stroke.

Example 22 — Two exotic powers added

Differentiate $x^{x^2 - 3} + (x - 3)^{x^2}$, for $x > 3$, w.r.t. $x$.

Step 1 — log each term separately: with $u = x^{x^2-3}$: $\log u = (x^2 - 3)\log x$, so $u' = u\left[\frac{x^2 - 3}{x} + 2x\log x\right]$.

Step 2 — the second term: with $v = (x-3)^{x^2}$: $\log v = x^2\log(x - 3)$, so $v' = v\left[\frac{x^2}{x - 3} + 2x\log(x - 3)\right]$.

Answer: $$\frac{dy}{dx} = x^{x^2-3}\left[\frac{x^2 - 3}{x} + 2x\log x\right] + (x-3)^{x^2}\left[\frac{x^2}{x-3} + 2x\log(x-3)\right]$$ — a sum of exotic powers is handled term by term, each with its own logarithm.

Example 23 — Modulus cubed is twice differentiable

If $f(x) = \vert x \vert^3$, show that $f''(x)$ exists for all real $x$ and find it.

Step 1 — split by sign: $f(x) = x^3$ for $x \geq 0$ and $f(x) = -x^3$ for $x < 0$.

Step 2 — differentiate each side twice: $f''(x) = 6x$ for $x > 0$ and $f''(x) = -6x$ for $x < 0$.

Step 3 — glue at $0$: $f'(x) = \pm 3x^2 \to 0$ from both sides (so $f'(0) = 0$ exists), and likewise both second-derivative formulas approach $0$ — the defining limits at $0$ agree.

Answer: $f''(x) = 6\vert x \vert$ for all real $x$. ∎ Cubing smooths the modulus corner enough for two derivatives — the third fails at $0$.

Example 24 — Differentiation as a formula factory

Using $\sin(A + B) = \sin A\cos B + \cos A\sin B$ and differentiation, obtain the sum formula for cosines.

Step 1 — treat $A$ as variable, $B$ as constant: differentiate both sides w.r.t. $A$.

Step 2 — left side: $\cos(A + B)$ by the chain rule (inner derivative $1$).

Step 3 — right side: $\cos A\cos B - \sin A\sin B$.

Answer: $\cos(A + B) = \cos A\cos B - \sin A\sin B$. ∎ Differentiating a trig identity yields its partner free of charge.

Example 25 — Exactly two bad points

Does there exist a function which is continuous everywhere but not differentiable at exactly two points?

Step 1 — build from corners: each $\vert x - a \vert$ is continuous everywhere and non-differentiable only at $a$, and sums of continuous functions are continuous.

Step 2 — the candidate: $f(x) = \vert x \vert + \vert x - 1 \vert$ — continuous everywhere; at $x = 0$ and $x = 1$ one summand has a corner while the other is locally linear, so the corners survive; everywhere else both pieces are smooth.

Answer: yes — $f(x) = \vert x \vert + \vert x - 1 \vert$ works, with non-differentiability at exactly $x = 0$ and $x = 1$.