The derivative of a function $f(x)$ with respect to x at a point x is defined by the limit: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ This formula represents the slope of the tangent line to the curve at the point x.

To find the derivative of $\sin(x)$ using first principles, we use the definition of the derivative: $\frac{d}{dx} \sin(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin(x)}{h}$

Step 1: Use the Sine Addition Formula:

The sine addition formula states: $\sin(x+h) = \sin(x)\cos(h) + \cos(x)\sin(h)$ Substitute this into the limit definition: $\frac{d}{dx} \sin(x) = \lim_{h \to 0} \frac{\sin(x)\cos(h) + \cos(x)\sin(h) - \sin(x)}{h}$

Step 2: Factor and Simplify:

Factor out $\sin(x)$ from the terms involving \sin(x): $\frac{d}{dx} \sin(x) = \lim_{h \to 0} \frac{\sin(x)[\cos(h) - 1] + \cos(x)\sin(h)}{h}$

This expression can be separated into two limits: $\frac{d}{dx} \sin(x) = \sin(x) \cdot \lim_{h \to 0} \frac{\cos(h) - 1}{h} + \cos(x) \cdot \lim_{h \to 0} \frac{\sin(h)}{h}$

Step 3: Evaluate the Limits:

There are two important trigonometric limits to know:

• $\lim_{h \to 0} \frac{\sin(h)}{h} = 1$
• $\lim_{h \to 0} \frac{\cos(h) - 1}{h} = 0$ Using these limits: $\frac{d}{dx} \sin(x) = \sin(x) \cdot 0 + \cos(x) \cdot 1 = \cos(x)$

So, the derivative of $\sin(x)$ is: $\frac{d}{dx} \sin(x) = \cos(x)$

To find the derivative of $\cos(x)$ using first principles, follow a similar process: $\frac{d}{dx} \cos(x) = \lim_{h \to 0} \frac{\cos(x+h) - \cos(x)}{h}$

Step 1: Use the Cosine Addition Formula:

The cosine addition formula states: $\cos(x+h) = \cos(x)\cos(h) - \sin(x)\sin(h)$ Substitute this into the limit definition: $\frac{d}{dx} \cos(x) = \lim_{h \to 0} \frac{\cos(x)\cos(h) - \sin(x)\sin(h) - \cos(x)}{h}$

Step 2: Factor and Simplify:

Factor out $\cos(x)$: $\frac{d}{dx} \cos(x) = \lim_{h \to 0} \frac{\cos(x)[\cos(h) - 1] - \sin(x)\sin(h)}{h}$

This can be separated into two limits: $\frac{d}{dx} \cos(x) = \cos(x) \cdot \lim_{h \to 0} \frac{\cos(h) - 1}{h} - \sin(x) \cdot \lim_{h \to 0} \frac{\sin(h)}{h}$

Step 3: Evaluate the Limits:

Using the same trigonometric limits: $\frac{d}{dx} \cos(x) = \cos(x) \cdot 0 - \sin(x) \cdot 1 = -\sin(x)$

So, the derivative of $\cos(x)$ is: $\frac{d}{dx} \cos(x) = -\sin(x)$

1. $\frac{d}{dx} (\sin x) = \cos x$ Proof:

• Expand using compound angle formula:

• Small Angle Approximations:

2. $\frac{d}{dx} (\cos x) = -\sin x$ Proof:

• Expand using compound angle formula:

Let $y = \tan x = \dfrac{\sin x}{\cos x}$

Using the quotient rule:

Proof:

Let $y = \sec x = (\cos x)^{-1}$

Using the chain rule:

e.g. If $y = \csc(\ln x + 4x), find\ \frac{dy}{dx}$. Using the chain rule:

(Differentiate expression without looking in bracket $\times$ by diff of bracket)

e.g. Find the differential of $y = \dfrac{\tan x}{\sin(e^x)}$. Using the quotient rule:

Find $f'(x)$ for $f(x) = \csc(2x+3) \cot(5x)$.

Example: Find $\dfrac{d}{dx} (\cos^2 x)$ 3. Rewrite $\cos^2 x \ as \ (\cos x)^2$

4. Use Chain Rule

[OCR 4753, Jan 2010, Q8

The curve crosses the x-axis at O, P and Q.

(i) Find the exact coordinates of P and Q.

(ii) Find the exact gradient of the curve at the point P.

Show also that the turning points of the curve occur when $x \tan 3x = \frac{1}{3}$.

(ii)

Let $x = \dfrac{\pi}{6} \Rightarrow \dfrac{dy}{dx} = \cos\left(\cancel{\frac{3\pi}{6}}\right) - \frac{3\pi}{6} \sin\left(\cancel{\frac{3\pi}{6}}\right) = -\dfrac {\pi}{2}$

$\cos\left(\cancel{\frac{3\pi}{6}}\right) = 0 \quad \sin\left(\cancel{\frac{3\pi}{6}}\right)=1$

Turning points occur when $\frac{dy}{dx} = 0 \Rightarrow \cos(3x) - 3x\sin(3x) = 0$