Question

Differentiate y=sin x using first principles

Answer 1

`\frac{\text{d}y}{\text{d}x}=\cos(x)`

Explanation:

Differentiating from First Principles is a technique to find an algebraic expression for the gradient at a particular point on the curve.

`\boxed{\begin{minipage}{5.6 cm}\underline{Differentiating from First Principles}\\\\\\$\text{f}\:'(x)=\displaystyle \lim_(h \to 0) \left[\frac{\text{f}(x+h)-\text{f}(x)}{(x+h)-x}\right]$\\\\\end{minipage}}`

The point (x + h, f(x + h)) is a small distance along the curve from (x, f(x)).

As h gets smaller, the distance between the two points gets smaller.

The closer the points, the closer the line joining them will be to the tangent line.

To differentiate y = sin(x) using first principles, substitute f(x + h) = sin(x + h) and f(x) = sin(x) into the formula:

`\implies \displaystyle \frac{\text{d}y}{\text{d}x}=\lim_(h \to 0) \left[(\sin(x+h)-\sin(x))/((x+h)-x)\right]`

Use the sin addition formula to expand sin(x + h).

`\boxed{\sin(A + B) = \sin A \cos B + \cos A \sin B}`

`\implies \displaystyle \frac{\text{d}y}{\text{d}x}=\lim_(h \to 0) \left[(\sin(x)\cos(h)+\cos(x)\sin(h)-\sin(x))/((x+h)-x)\right]`

`\implies \displaystyle \frac{\text{d}y}{\text{d}x}=\lim_(h \to 0) \left[(\sin(x)\cos(h)-\sin(x)+\cos(x)\sin(h))/(h)\right]`

`\implies \displaystyle \frac{\text{d}y}{\text{d}x}=\lim_(h \to 0) \left[(\sin(x)(\cos(h)-1)+\cos(x)\sin(h))/(h)\right]`

Separate the sin(x) and cos(x) terms into two fractions:

`\implies \displaystyle \frac{\text{d}y}{\text{d}x}=\lim_(h \to 0) \left[(\sin(x)(\cos(h)-1))/(h)+(\cos(x)\sin(h))/(h)\right]`

When h gets really small, we can use the small angle approximation to rewrite cos(h).

`\boxed{\cos (h) \approx 1-(1)/(2)h^2}`

`\implies \displaystyle \frac{\text{d}y}{\text{d}x}=\lim_(h \to 0) \left[(\sin(x)\left(1-(1)/(2)h^2-1\right))/(h)+(\cos(x)\cdot h)/(h)\right]`

`\implies \displaystyle \frac{\text{d}y}{\text{d}x}=\lim_(h \to 0) \left[(\sin(x)\left(-(1)/(2)h^2\right))/(h)+(\cos(x)\cdot h)/(h)\right]`

Cancel the common factor, h:

`\implies \displaystyle \frac{\text{d}y}{\text{d}x}=\lim_(h \to 0) \left[-(1)/(2)h\sin(x)+\cos(x)\right]`

As h → 0, the first term → 0:

`\implies \displaystyle \frac{\text{d}y}{\text{d}x}=\cos(x)`