Differentiating trigonometric functions

In a nutshell

Standard trigonometric derivatives can be derived by applying the differentiation rules you already know to trigonometric identities.

Sine and cosine recap

Recall that there is a cyclic property to the sequence formed by differentiating the sine and cosine functions. In particular, by differentiating each term, you obtain the next in the sequence below:

​$$...\rightarrow\sin(x)\rightarrow\cos(x)\rightarrow-\sin(x)\rightarrow-\cos(x)\rightarrow\sin(x)\rightarrow...$$​​

Using identities

By recognising $$\tan(x)$$ as $$\frac{\sin(x)}{\cos(x)}$$, the quotient rule can be used. Recall the quotient rule says that for $$f(x)=\frac{u(x)}{v(x)}$$, the first derivative is $$f'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{v(x)^2}$$. Hence let:

​$$u(x)=\sin(x)$$ and $$v(x)=\cos(x)$$

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Consequently, by differentiating these with respect to $$x$$, you have:

​$$u'(x)=\cos(x)$$ and $$v'(x)=-\sin(x)$$​​

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Thus:

​$$\begin{aligned}f(x)=\tan(x)&\rightarrow f'(x)=\frac{\cos^2(x)+\sin^2(x)}{\cos(x)^2}\\&\rightarrow \boxed{f'(x)=\frac1{\cos^2(x)}=\sec^2(x)}\end{aligned}$$​​

In a very similar way you can show, for a constant $$k$$, that $$\tan(kx)$$ differentiates to $$k\sec^2(kx)$$.

Example 1

Differentiate $$f(x)=\sec(x)$$ with respect to $$x$$.

Again, use the quotient rule with $$u(x)=1$$ and $$v(x)=\cos(x)$$. It follows that:

$$\begin{aligned}f(x)=\sec(x)&\rightarrow f'(x)=\frac{0+\sin(x)}{\cos(x)^2}\\&\rightarrow f'(x)=\frac{\sin(x)}{\cos^2(x)}=\underline{\tan(x)\sec(x)}\end{aligned}$$​​

Using a similar process, you can show the following:

​$$\begin{aligned}f(x)=\cosec(kx)&\rightarrow \boxed{f'(x)=-k\cosec(kx)\cot(kx)}\\f(x)=\sec(kx)&\rightarrow \boxed{f'(x)=k\sec(kx)\tan(kx)}\\f(x)=\cot(kx)&\rightarrow \boxed{f'(x)=-k\cosec^2(kx)}\end{aligned}$$​​

Example 2

Differentiate $$g(x)={3x^5}{\cot(4x)}$$ with respect to $$x$$. 

Here, you have a product of the functions $$u(x)=3x^5$$ and $$v(x)=\cot(4x)$$. The product rule says that for $$g(x)=u(x)v(x)\rightarrow g'(x)=u'(x)v(x)+u(x)v'(x)$$. 

So first find $$u'(x)$$ and $$v'(x)$$:

$$u'(x)=15x^4$$ and $$v'(x)=-4\cosec^2(4x)$$​​

Therefore:

$$g'(x)=\underline{15x^4\cot(4x)-12x^5\cosec^2(4x)}$$​​

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