Derivatives of standard functions and applications

Combining the standard derivatives table with the rules of differentiation, you can differentiate almost any expression in this chapter. Then a few applications make the calculus pay off: finding tangent lines, rates of change, and simple optimisation.

Quick recap

Combine with R1–R5 from the previous subtopic.

Deriving the trig derivatives

We proved $(\sin x)' = \cos x$ from first principles. Similarly, $(\cos x)' = -\sin x$.

For $\tan x = \sin x / \cos x$, apply the quotient rule: $(\tan x)' = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \sec^2 x.$

Similarly for $\cot, \sec, \csc$.

Tangent line

The tangent to $y = f(x)$ at $x = a$ has slope $f'(a)$ and passes through $(a, f(a))$. Equation: $y - f(a) = f'(a)(x - a).$

The normal line is perpendicular to the tangent; its slope is $-1/f'(a)$.

Rate of change

If $y = f(x)$ and $x$ changes at a known rate, then $y$ changes at rate $\dfrac{dy}{dx} \cdot \dfrac{dx}{dt}$ (chain rule, formalised in Class XII; here we mostly stay with $dy/dx$).

Worked examples

Example 1. Find the equation of the tangent to $y = x^2 - 3 x + 2$ at $x = 1$.

$f(1) = 1 - 3 + 2 = 0$. $f'(x) = 2 x - 3$, so $f'(1) = -1$. Tangent: $y - 0 = -1(x - 1) \Rightarrow y = -x + 1$.

Example 2. Find the points on $y = x^3 - 3 x$ where the tangent is horizontal.

$f'(x) = 3 x^2 - 3 = 0 \Rightarrow x = \pm 1$. Points: $(1, -2)$ and $(-1, 2)$.

Example 3. Differentiate $f(x) = e^x \sin x$.

By product rule: $f'(x) = e^x \sin x + e^x \cos x = e^x (\sin x + \cos x)$.

Example 4. Differentiate $f(x) = \dfrac{x^2 + 1}{x + 2}$.

By quotient rule: $f'(x) = \dfrac{2 x (x + 2) - (x^2 + 1)}{(x + 2)^2} = \dfrac{x^2 + 4 x - 1}{(x + 2)^2}$.

Example 5. Find $\dfrac{d}{dx}\left[\dfrac{1}{1 + \tan x}\right]$.

Quotient rule: $\dfrac{0 \cdot (1 + \tan x) - 1 \cdot \sec^2 x}{(1 + \tan x)^2} = \dfrac{-\sec^2 x}{(1 + \tan x)^2}$.

Some computations beyond the basics

Logarithmic differentiation. For $y = x^x$ (here $x > 0$): take $\ln$. $\ln y = x \ln x$. Differentiate (implicitly, a hint of Class XII): $\dfrac{y'}{y} = \ln x + 1$, so $y' = x^x (\ln x + 1)$.

This trick is sometimes useful even in Class XI for products of many factors.

Try it yourself

1. Differentiate $f(x) = x^4 - 2 x^3 + 7 x - 5$.
2. Differentiate $f(x) = 3 \sin x - 2 \cos x$.
3. Differentiate $f(x) = x \tan x$.
4. Find the tangent to $y = x^2$ at $x = 3$.
5. Find points on $y = x^3 - 6 x^2 + 9 x + 1$ where the tangent is horizontal.
6. Differentiate $f(x) = \dfrac{\sin x}{x}$.
7. Differentiate $f(x) = (x + 1) \sin x$.
8. Differentiate $f(x) = \dfrac{1 - \sin x}{1 + \sin x}$.
9. Find $f'(0)$ for $f(x) = x e^x$.
10. Find the slope of the tangent to $y = \ln x$ at $x = e$.
11. Find the normal line to $y = x^2$ at $x = 1$.
12. A particle has position $s(t) = t^3 - 6 t^2 + 9 t$. Find the velocity and acceleration at $t = 2$. ($v = s'$, $a = v'$.)

Pitfalls / Tricks

• Use the right rule: product for $u v$, quotient for $u/v$.
• Memorise the standard derivatives table.
• For tangent line problems, plug in $x = a$ after differentiating.
• Insight. Differentiation is mechanical: identify the structure (sum / product / quotient), apply the corresponding rule, and use standard derivatives. Once you internalise this, almost every Class XI derivative is a matter of seconds.