Derivatives of Standard Functions
Trigonometric Functions
The derivatives of the basic trigonometric functions are fundamental in calculus:
- $\frac{d}{dx}(\sin x) = \cos x$
- $\frac{d}{dx}(\cos x) = -\sin x$
- $\frac{d}{dx}(\tan x) = \sec^2 x$
Let's differentiate $f(x) = \sin(2x)$:
Using the chain rule (which we'll explore later), we get: $f'(x) = \cos(2x) \cdot 2 = 2\cos(2x)$
NoteThe derivatives of inverse trigonometric functions, while not explicitly mentioned in the syllabus, are often useful in more advanced problems:
- $\frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1-x^2}}$
- $\frac{d}{dx}(\arccos x) = -\frac{1}{\sqrt{1-x^2}}$
- $\frac{d}{dx}(\arctan x) = \frac{1}{1+x^2}$
Exponential and Logarithmic Functions
The exponential and natural logarithm functions have particularly elegant derivatives:
- $\frac{d}{dx}(e^x) = e^x$
- $\frac{d}{dx}(\ln x) = \frac{1}{x}$
Differentiate $g(x) = e^{3x}$:
Again, using the chain rule: $g'(x) = e^{3x} \cdot 3 = 3e^{3x}$
Power Function
For the general power function:
- $\frac{d}{dx}(x^n) = nx^{n-1}$, where $n$ is a rational number
This rule is incredibly versatile, applying to integer, fractional, and negative exponents.
ExampleDifferentiate $h(x) = x^{-1/2}$:
$h'(x) = -\frac{1}{2}x^{-3/2}$
Common MistakeStudents often forget that this rule applies to negative and fractional exponents. Remember, $x^{-1} = \frac{1}{x}$ and $x^{1/2} = \sqrt{x}$.
Chain Rule
The chain rule is a powerful tool for differentiating composite functions. If $y = f(u)$ and $u = g(x)$, then:
$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$
In Leibniz notation, this is often written as:
$$\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$$
ExampleDifferentiate $y = \sin(x^2)$:
Let $u = x^2$, then $y = \sin(u)$
$\frac{dy}{du} = \cos(u)$ $\frac{du}{dx} = 2x$
Therefore, $\frac{dy}{dx} = \cos(x^2) \cdot 2x = 2x\cos(x^2)$
TipWhen applying the chain rule, it can be helpful to think of it as "inside function" times "outside function". The derivative of the inside function is multiplied by the derivative of the outside function composed with the inside function.
Product Rule
The product rule is used to differentiate the product of two functions. If $y = u(x)v(x)$, then:
$$\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$$
Or in more compact notation:
$$(uv)' = u'v + uv'$$
ExampleDifferentiate $y = x\sin(x)$:
Let $u = x$ and $v = \sin(x)$
$\frac{dy}{dx} = x\cos(x) + \sin(x)$
NoteThe product rule can be extended to more than two functions. For three functions $u$, $v$, and $w$:
$(uvw)' = u'vw + uv'w + uvw'$
Quotient Rule
The quotient rule is used to differentiate the quotient of two functions. If $y = \frac{u(x)}{v(x)}$, then:
$$\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$$
Or in more compact notation:
$$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$$
ExampleDifferentiate $y = \frac{\cos(x)}{x}$:
Let $u = \cos(x)$ and $v = x$
$\frac{dy}{dx} = \frac{x(-\sin(x)) - \cos(x)(1)}{x^2} = \frac{-x\sin(x) - \cos(x)}{x^2}$
TipA helpful mnemonic for remembering the quotient rule is "low d-high minus high d-low, over low-low".
Related Rates of Change
Related rates problems involve finding the rate of change of one quantity with respect to another when the two quantities are related by an equation and are both changing with respect to time.
The key steps in solving related rates problems are:
- Identify the known and unknown rates of change
- Write an equation relating the variables
- Differentiate both sides of the equation with respect to time
- Substitute known values and solve for the unknown rate
A spherical balloon is being inflated at a rate of 10 cm³/s. How fast is the radius increasing when the radius is 5 cm?
The volume of a sphere is $V = \frac{4}{3}\pi r^3$
Differentiating with respect to time:
$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$
We know $\frac{dV}{dt} = 10$ cm³/s and $r = 5$ cm. Substituting:
$10 = 4\pi(5^2)\frac{dr}{dt}$
Solving for $\frac{dr}{dt}$:
$\frac{dr}{dt} = \frac{10}{100\pi} \approx 0.0318$ cm/s
Common MistakeIn related rates problems, students often forget to differentiate with respect to time. Remember, even if a variable like radius ($r$) is given as a constant at a specific moment, it's still changing with respect to time, so $\frac{dr}{dt}$ is not zero.
Connection to Optimization and Extrema
The techniques of differentiation are crucial in finding maximum and minimum points (extrema) of functions, which is a key component of optimization problems.
- The first derivative test: If $f'(c) = 0$ or $f'(c)$ does not exist, then $c$ is a critical point of $f$.
- The second derivative test: If $f'(c) = 0$ and $f''(c) > 0$, then $f$ has a local minimum at $c$. If $f''(c)
< 0$, then $f$ has a local maximum at $c$.
ExampleFind the maximum value of $f(x) = x^3 - 3x^2 + 1$ on the interval $[0,3]$.
$f'(x) = 3x^2 - 6x$ $f'(x) = 3x(x-2)$
Critical points: $x = 0$ and $x = 2$
Evaluating $f$ at the critical points and endpoints: $f(0) = 1$ $f(2) = -3$ $f(3) = 1$
The maximum value is 1, occurring at $x = 0$ and $x = 3$.
NoteOptimization problems often involve constraints. In such cases, you may need to use techniques like substitution or Lagrange multipliers (in more advanced courses) to solve the problem.
