The derivative of trigonometric functions

Differentiation: The derivative of standard functions

The derivative of trigonometric functions

There are rules to determine the derivative of the trigonometric functions #\blue{\sin}(x), \green{\cos}(x)# and #\purple{\tan}(x)#. These rules only apply to trigonometric functions in radians.

The derivative of sine

\[\dfrac{\dd}{\dd x}\blue{\sin}(x)=\green{\cos}(x)\]

Example

\[\dfrac{\dd}{\dd x}(3\cdot\blue{\sin}(x))=3\cdot\green{\cos}(x)\]

The derivative of cosine

\[\dfrac{\dd}{\dd x}\green{\cos}(x)=-\blue{\sin}(x)\]

Example

\[\dfrac{\dd}{\dd x}(4\cdot\green{\cos}(x))=-4\cdot\blue{\sin}(x)\]

We can write the derivative of tangent in two ways.

The derivative of tangent

\[\begin{array}{rcl}\dfrac{\dd}{\dd x}\purple{\tan}(x)&=&\dfrac{1}{\green{\cos}(x) ^2}\\\dfrac{\dd}{\dd x}\purple{\tan}(x)&=&1+\purple{\tan} (x)^2\end{array}\]

Example

\[\begin{array}{rcl}\dfrac{\dd}{\dd x}(3\cdot\purple{\tan}(x))&=&\dfrac{3}{\green{\cos}(x)^2}\\\dfrac{\dd}{\dd x}(3\cdot\purple{\tan}(x))&=&3+3\cdot\purple{\tan}(x)^2\end{array}\]

In the chapter on trigonometric functions we have seen that #\tan(x)# is defined as #\frac{\sin(x)}{\cos(x)}#. Now that we know the derivatives of #\sin(x)# and #\cos(x)#, we can determine the derivative of #\tan(x)# using the quotient rule, which we will talk about later.

An alternative proof makes use of the product rule and the chain rule. We write \[f(x)=\tan(x) = \dfrac{\sin(x)}{\cos(x)}=\sin(x)\cdot \dfrac{1}{\cos(x)}\] Product rule gives \[f'(x)=\cos(x)\cdot \dfrac{1}{\cos(x)} + \sin(x)\cdot \left(\dfrac{1}{\cos(x)}\right)'=1+\sin(x)\cdot \left(\dfrac{1}{\cos(x)}\right)'\] We can calculate the derivative #\left(\dfrac{1}{\cos(x)}\right)'# by writing #\dfrac{1}{\cos(x)}# as the composition of #\dfrac{1}{x}# and #\cos(x)#. Applying the chain rule gives \[\left(\dfrac{1}{\cos(x)}\right)'= -\dfrac{1}{\cos(x)^2}\cdot (-\sin(x))=\dfrac{\sin(x)}{\cos(x)^2}\] We now conclude the proof: \[f'(x)=1 + \sin(x)\cdot \left(\dfrac{1}{\cos(x)}\right)'=1+\sin(x)\cdot \dfrac{\sin(x)}{\cos(x)^2}=1+\dfrac{\sin(x)^2}{\cos(x)^2}=1+\tan(x)^2\]

It is easy to see that both expressions are the same. We defined #\tan(x)# as \[\tan(x)=\dfrac{\sin(x)}{\cos(x)}\] Squaring gives \[\tan(x)^2=\dfrac{\sin(x)^2}{\cos(x)^2}\] By making use of the identity #\sin(x)^2+\cos(x)^2=1#, we can rewrite this to \[\tan(x)^2=\dfrac{1-\cos(x)^2}{\cos(x)^2}=\dfrac{1}{\cos(x)^2}-1\] We add #1# to both sides to get the equality: \[1+\tan(x)^2=\dfrac{1}{\cos(x)^2}\]

To calculate the derivatives of trigonometric functions, we often need the product rule and sum rule. We can also use the chain rule for finding the derivatives of trigonometric functions.

Determine the derivative of the function #2-6\cdot \cos \left(x\right)#.

#\frac{\dd}{\dd x} \left(2-6\cdot \cos \left(x\right)\right) =# # 6\cdot \sin \left(x\right)#

\[\begin{array}{rcl}
\dfrac{\dd}{\dd x} \left(2-6\cdot \cos \left(x\right)\right) &=&\displaystyle \frac{\dd}{\dd x} \left(-6 \cdot\cos(x)\right) +\frac{\dd}{\dd x} \left(2\right) \\
&&\phantom{xxx}\blue{\text{sum rule}}\\
&=& \displaystyle-6 \cdot \frac{\dd}{\dd x} \left(\cos(x)\right) +0 \\
&&\phantom{xxx}\blue{\text{constant rule and derivative of constant is }0}\\
&=& \displaystyle6\cdot \sin \left(x\right) \\
&&\phantom{xxx}\blue{\text{the derivative of }\cos(x)\text{ is }-\sin \left(x\right)}\\
\end{array}\]

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