Proving the derivative of sin x

The limit of sin h over h as h tends to zero, proved geometrically, and the proof from first principles that the derivative of sin x is cos x, with the derivatives of cos x and tan x that follow.

The key limit

sin h ÷ h approaches 1

Drag the point towards 0. The value gets as close to 1 as you like, although h = 0 itself is undefined.

Why it is true●●●

Take a unit circle and a small angle $h$ at the centre, with $0 < h < \tfrac{\pi}{2}$.

The triangle inside the sector has area $\tfrac12\sin h$. The sector has area $\tfrac12 h$. The larger right triangle reaching the tangent line has area $\tfrac12\tan h$.

Three areas

The regions are nested, so $\sin h < h < \tan h$.

Compare

$1 < \dfrac{h}{\sin h} < \dfrac{1}{\cos h}$, so $\cos h < \dfrac{\sin h}{h} < 1$.

Divide by sin h

As $h \to 0$, $\cos h \to 1$, so $\dfrac{\sin h}{h}$ is squeezed to $1$.

Squeeze

The derivative of sin x

Proof from first principles●●●

Show that $\dfrac{d}{dx}\sin x = \cos x$.

$\sin(x + h) - \sin x = 2\cos\left(x + \tfrac{h}{2}\right)\sin\tfrac{h}{2}$.

Sums to products

$\dfrac{\sin(x + h) - \sin x}{h} = \cos\left(x + \tfrac{h}{2}\right) \times \dfrac{\sin\frac{h}{2}}{\frac{h}{2}}$.

Divide by h

As $h \to 0$, the first factor tends to $\cos x$ and the second to $1$. So $\dfrac{d}{dx}\sin x = \cos x$.

Take the limit

Cosine and tangent follow

$\cos x = \sin\left(\tfrac{\pi}{2} - x\right)$, so $\dfrac{d}{dx}\cos x = -\cos\left(\tfrac{\pi}{2} - x\right) = -\sin x$.
$\tan x = \dfrac{\sin x}{\cos x}$, so by the quotient rule $\dfrac{d}{dx}\tan x = \dfrac{\cos^2 x + \sin^2 x}{\cos^2 x} = \sec^2 x$.

Flashcards

$displaystylelim_{h o 0}rac{sin h}{h}$

$1$, with $h$ in radians.

The inequality in the proof

$\sin h < h < \tan h$

Derivative of $ an x$ by the quotient rule

$\dfrac{\cos^2 x + \sin^2 x}{\cos^2 x} = \sec^2 x$

Practice

Find $\displaystyle\lim_{h \to 0}\frac{\sin h}{h}$.
Find $\displaystyle\lim_{h \to 0}\frac{\sin 3h}{h}$.
Find $\displaystyle\lim_{h \to 0}\frac{\sin 5h}{\sin 2h}$.
Find $\displaystyle\lim_{h \to 0}\frac{\tan h}{h}$.
The geometric proof uses the inequality:
$\sin(x + h) - \sin x = $
The derivative of $\cos x$ follows from writing $\cos x$ as:
Why must $x$ be in radians?
$\dfrac{d}{dx}\tan x = $