HSC Maths Advanced Syllabus

How to Differentiate Trigonometric Functions

The simple derivatives of the trigonometric functions are as follows:

$$\frac{d}{dx}sinx=cosx \qquad\quad\frac{d}{dx}cosx=-sinx \qquad\quad\frac{d}{dx}tanx=sec^2x$$

When trigonometric functions are represented in the form of $ax +b$ their derivatives are as follows:

$$\frac{d}{dx}sin(ax+b)=a\times cos(ax+b)$$ $$\frac{d}{dx}cos(ax+b)=-a\times sin(ax+b)$$ $$\frac{d}{dx}tan(ax+b)=a\times sec^2(ax+b)$$

Using the Chain Rule With Trigonometric Functions

As a quick review the chain rule is noted as:

$$\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}$$

Although the chain rule can be used for simpler trigonometric functions such as $y=cos(2x+1)$, it is easier to apply the formulas given above. However, once functions become increasingly complicated, chain rule can be applied to find the derivative.

For example, when given the function $y=sin(5x^2+1)$, we are unable to apply any of the formulas given above. Thus, we use chain rule to find its derivative.

First we must define the variable $u$ and derive it :

$$u=5x^2 +1\\\frac{du}{dx}=2\times 5x\\=10x$$

Now, substituting $u$ into the original equation and deriving it, we can find $\frac{dy}{du}$

$$y=sin(u)\\\frac{dy}{du}=u^{'}cos(u)$$

Substituting $u$ and its derivative into $y$, we can determine $\frac{dy}{dx}$

$$\frac{dy}{dx}=10xcos(5x^2+1)$$