The derivative of a function $f$ at a point $x=a$ is the limit of the difference quotient:$$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$$An equivalent form: $f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$.
Geometric meaning: $f'(a)$ is the slope of the tangent line to the graph of $y=f(x)$ at the point $(a,\,f(a))$. Physical meaning: the instantaneous rate of change of $f$ at $x=a$.
On the YOS exam you must recognize limits that hide a derivative. Useful patterns:$$\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}=f'(a),\qquad \lim_{h\to 0}\frac{f(a+kh)-f(a)}{h}=k\,f'(a),$$$$\lim_{h\to 0}\frac{f(a+h)-f(a-h)}{2h}=f'(a).$$So if a problem gives such a limit, compute $f'(x)$ by the rules below and substitute $x=a$ — do not evaluate the limit directly.
Power and constant:$$(c)'=0,\qquad (x^n)'=nx^{n-1},\qquad (\sqrt{x})'=\frac{1}{2\sqrt{x}},\qquad \left(\frac{1}{x}\right)'=-\frac{1}{x^2}$$Exponential and logarithmic:$$(e^x)'=e^x,\qquad (a^x)'=a^x\ln a,\qquad (\ln x)'=\frac{1}{x},\qquad (\log_a x)'=\frac{1}{x\ln a}$$Trigonometric:$$(\sin x)'=\cos x,\qquad (\cos x)'=-\sin x,\qquad (\tan x)'=\frac{1}{\cos^2 x}=1+\tan^2 x,\qquad (\cot x)'=-\frac{1}{\sin^2 x}$$Inverse trigonometric:$$(\arcsin x)'=\frac{1}{\sqrt{1-x^2}},\qquad (\arccos x)'=-\frac{1}{\sqrt{1-x^2}},\qquad (\arctan x)'=\frac{1}{1+x^2}$$Memorize this table: every exam problem reduces to it after applying the rules of differentiation.
Linearity:$$(c\,f)'=c\,f',\qquad (f\pm g)'=f'\pm g'$$Product rule:$$(f\cdot g)'=f'g+fg'$$Quotient rule:$$\left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^2},\qquad g(x)\neq 0$$Typical exam use of the product rule: if $f(x)=(x^2-1)\,g(x)$, then$$f'(x)=2x\,g(x)+(x^2-1)\,g'(x).$$To find $f'(2)$ you only need the values $g(2)$ and $g'(2)$ given in the problem: $f'(2)=4g(2)+3g'(2)$.
For a composite function:$$\big(f(g(x))\big)'=f'(g(x))\cdot g'(x)$$Differentiate the outer function keeping the inner one unchanged, then multiply by the derivative of the inner function.
Exam-style examples:$$\big((x^2-5x+7)^3\big)'=3(x^2-5x+7)^2\cdot(2x-5)$$$$\big(\sin(4x+3)\big)'=4\cos(4x+3),\qquad (\tan 5x)'=\frac{5}{\cos^2 5x}$$$$\big(\arccos(x^3)\big)'=-\frac{3x^2}{\sqrt{1-x^6}},\qquad \big(\arcsin\sqrt{x}\big)'=\frac{1}{2\sqrt{x}\sqrt{1-x}}$$The chain rule also handles "implicit composition" problems: if $f(g(x))=h(x)$, differentiate both sides to get $f'(g(x))\cdot g'(x)=h'(x)$, then choose a convenient value of $x$.
Combining the table with the chain rule ($u=u(x)$):$$\big(a^{u}\big)'=a^{u}\ln a\cdot u',\qquad \big(e^{u}\big)'=e^{u}\cdot u',\qquad \big(\ln u\big)'=\frac{u'}{u}$$Examples:$$\big(3^{\cos x}\big)'=3^{\cos x}\ln 3\cdot(-\sin x)=-3^{\cos x}\sin x\ln 3$$$$\big(\ln(\sin x)\big)'=\frac{\cos x}{\sin x}=\cot x$$Absolute value: for $u(x)\neq 0$$$\big(|u|\big)'=\frac{u}{|u|}\cdot u'$$In practice: determine the sign of $u$ near the given point and remove the modulus. For $f(x)=|x^3-8|$ near $x=-2$ we have $x^3-8<0$, so $f(x)=8-x^3$ and $f'(x)=-3x^2$, hence $f'(-2)=-12$. At a point where $u=0$ and $u'\neq 0$ the function $|u|$ is not differentiable (corner point).
The one-sided derivatives at $x=a$ are$$f'(a^-)=\lim_{h\to 0^-}\frac{f(a+h)-f(a)}{h},\qquad f'(a^+)=\lim_{h\to 0^+}\frac{f(a+h)-f(a)}{h}.$$A function is differentiable at $a$ if and only if it is continuous at $a$ and $f'(a^-)=f'(a^+)$.
For a piecewise function, differentiate each branch separately: $f'(a^-)$ is the derivative of the left branch evaluated at $a$, and $f'(a^+)$ is the derivative of the right branch at $a$ (provided $f$ is continuous at $a$).
Mini-example. Let $f(x)=x^2+1$ for $x\le 1$ and $f(x)=3x-1$ for $x>1$ (continuous at $x=1$, both equal $2$). Then$$f'(1^-)=2x\big|_{x=1}=2,\qquad f'(1^+)=3,\qquad f'(1^-)+f'(1^+)=5.$$Since $2\neq 3$, $f$ is not differentiable at $x=1$, but the sum of one-sided derivatives is still well defined — a frequent exam question.
The second derivative is the derivative of the derivative:$$f''(x)=\big(f'(x)\big)'$$To compute it, differentiate twice. For polynomials it is often easier to expand the brackets first.
Exam pattern with a parameter. Let $f(x)=(x+m)^2(x-1)$ and $f''(1)=16$. Expand:$$f(x)=x^3+(2m-1)x^2+(m^2-2m)x-m^2$$$$f'(x)=3x^2+2(2m-1)x+m^2-2m,\qquad f''(x)=6x+2(2m-1)$$Then $f''(1)=6+4m-2=4m+4=16$, so $m=3$.
The second derivative also describes concavity: $f''>0$ — the graph is concave up, $f''<0$ — concave down; points where $f''$ changes sign are inflection points.
The tangent line to the graph of $y=f(x)$ at the point $x=a$:$$y=f(a)+f'(a)(x-a)$$Here $f'(a)$ is the slope of the tangent; if the tangent makes an angle $\alpha$ with the positive $x$-axis, then $f'(a)=\tan\alpha$.
The normal line at the same point (perpendicular to the tangent, $f'(a)\neq 0$):$$y=f(a)-\frac{1}{f'(a)}(x-a)$$Typical exam questions: find the tangent at a given point; find the point where the tangent is parallel to a given line ($f'(a)$ equals the slope of that line); find where the tangent is horizontal ($f'(a)=0$).
On an interval where $f'(x)>0$ the function increases; where $f'(x)<0$ it decreases.
Critical points are points where $f'(x)=0$ or $f'$ does not exist. Local extrema of a differentiable function can occur only at critical points.
First derivative test: if at a critical point $x_0$ the sign of $f'$ changes from $+$ to $-$, then $x_0$ is a local maximum; from $-$ to $+$ — a local minimum; if the sign does not change, there is no extremum.
Second derivative test: if $f'(x_0)=0$ and $f''(x_0)>0$, then $x_0$ is a local minimum; if $f''(x_0)<0$ — a local maximum.
On a closed interval $[a,b]$ compare the values of $f$ at critical points inside the interval and at the endpoints to find the greatest and least values.
Some YOS problems give a relation between $f$ and $f'$, for example$$f(x)-f'(x)=4x^2+6x+7.$$Method: since differentiation lowers the degree by one, the right side forces $f$ to be a polynomial of the same degree. Set $f(x)=ax^2+bx+c$, then $f'(x)=2ax+b$ and$$f(x)-f'(x)=ax^2+(b-2a)x+(c-b)=4x^2+6x+7.$$Matching coefficients: $a=4$; $b-2a=6$ gives $b=14$; $c-b=7$ gives $c=21$. Hence$$f(x)=4x^2+14x+21,$$and any requested value follows, e.g. $f(0)=21$, $f(1)=39$.
The same idea (undetermined coefficients) works for relations like $f(x)+f'(x)=\ldots$ or $f'(x)-2f(x)=\ldots$ with a polynomial right-hand side.
Problem. Let $f(x)=ax^3+bx^2-3x-4$. If $f'(1)=3$ and $f'(-1)=-3$, find $a+b$.
Solution.Step 1. Differentiate: $f'(x)=3ax^2+2bx-3$.Step 2. Substitute $x=1$: $f'(1)=3a+2b-3=3$, so $3a+2b=6$.Step 3. Substitute $x=-1$: $f'(-1)=3a-2b-3=-3$, so $3a-2b=0$.Step 4. Add the equations: $6a=6$, hence $a=1$; then $2b=3a=3$, so $b=\frac{3}{2}$.Step 5. Therefore$$a+b=1+\frac{3}{2}=\frac{5}{2}.$$Answer: $\frac{5}{2}$.
Problem. Given $f(2x^2-3x)=x^4+x^2$ for all $x$, find $f'(-1)$.
Solution.Step 1. Differentiate both sides with respect to $x$ using the chain rule:$$f'(2x^2-3x)\cdot(4x-3)=4x^3+2x.$$Step 2. We need the inner expression to equal $-1$: solve $2x^2-3x=-1$, i.e. $2x^2-3x+1=0$, which gives $x=1$ (a convenient root).Step 3. Substitute $x=1$:$$f'(-1)\cdot(4\cdot 1-3)=4\cdot 1+2\cdot 1\quad\Rightarrow\quad f'(-1)\cdot 1=6.$$Answer: $f'(-1)=6$.
Key idea: never try to find the formula of $f$ itself — differentiate the identity and pick the value of $x$ that makes the inner function equal to the required point.
Problem. Let $f(x)=|x^3-8|$. Find $f'(-2)$.
Solution.Step 1. Determine the sign of the expression inside the modulus near $x=-2$:$$x^3-8\big|_{x=-2}=(-2)^3-8=-16<0.$$Since $x^3-8<0$ for all $x<2$, on a whole neighborhood of $x=-2$ we have$$f(x)=-(x^3-8)=8-x^3.$$Step 2. Differentiate this branch: $f'(x)=-3x^2$.Step 3. Substitute: $f'(-2)=-3\cdot(-2)^2=-12$.Answer: $f'(-2)=-12$.
Note: at $x=2$ the expression under the modulus vanishes and changes sign, so $f$ is not differentiable there (a corner). But $x=-2$ lies strictly inside the region where the sign is constant, so the derivative exists.
Problem. Find the local extremum of $f(x)=2x^3-6x+9$ on the interval $[-1,3]$.
Solution.Step 1. Differentiate: $f'(x)=6x^2-6=6(x-1)(x+1)$.Step 2. Critical points: $f'(x)=0$ at $x=1$ and $x=-1$. Inside the interval $(-1,3)$ only $x=1$ lies ($x=-1$ is an endpoint).Step 3. Sign of $f'$: on $(-1,1)$ we have $f'<0$ (decreasing), on $(1,3)$ we have $f'>0$ (increasing). The sign changes from $-$ to $+$, so $x=1$ is a local minimum.Step 4. The local minimum value:$$f(1)=2-6+9=5.$$Answer: local minimum $5$ at $x=1$.
For the greatest and least values on $[-1,3]$ also compare the endpoints: $f(-1)=13$, $f(3)=45$; so the least value on the interval is $5$ and the greatest is $45$.
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