Derivatives of Trigonometric Functions
Derivatives of trigonometric functions have applications ranging from electronics to computer programming and modeling different cyclic functions.
To find the derivative of \(\sin \theta,\) we can use the definition of the derivative
\[ f'(x) = \lim_{h \rightarrow 0} \frac{ f(x+h) - f(x) } { h } .\]
So for \(f(x)=\sin x,\)
\[f'(x) = \lim_{h \rightarrow 0} \frac{\sin(x+h)-\sin x}{h} .\]
Using the addition identity, we can expand
\[ \sin(x+h) = \sin x \cos h + \sin h \cos x.\]
Plugging this in our limit gives
\[\begin{align} f'(x) &= \lim_{h \rightarrow 0} \frac{\sin x \cos h + \sin h \cos x-\sin x}{h} \\\\ &= \lim_{h \rightarrow 0} \frac{\sin x (\cos h - 1) + \sin h \cos x}{h}. \end{align} \]
We can now separate this into two limits:
\[f'(x) = \sin x \times \lim_{h \rightarrow 0}\frac{\cos h-1}{h} + \cos x \times \lim_{h \rightarrow 0}\frac{\sin h}{h}. \]
We can see that by using Taylor series expansion, the first limit converges to \(0\) and the second limit converges to \(1\), which can also be shown in the following graphs:
im
im2
Separating and evaluating the limits, we get
\[f'(x) = \sin x \times 0+ \cos x \times 1 = \cos x.\]
What is the slope of the tangent line on the sine curve at the point \(x=\frac{\pi}{2}?\)
From the definition of the derivative, we know that the slope of a tangent line at a point on a curve is the derivative of its function at that point. Thus
\[ (\sin x )'\Big|_{x=\frac{\pi}2} = \cos\frac{\pi}{2} = 0 . \ _\square\]
Now that we know \(f'(x) = \cos x\) for \(f(x)=\sin x,\) we can easily find the derivative of \(f(x)=\cos x.\)
Remember the fundamental identity
\[\cos x = \sin\left(\frac{\pi}{2} - x\right). \]
Differentiating both sides and using the chain rule, we get
\[\begin{align} (\cos x)' &= \left(\sin\Big(\frac{\pi}{2} - x\Big)\right)' \times \left(\frac{\pi}{2} - x\right) ' \\ &= \cos\left(\frac{\pi}{2} - x\right) \times (- 1)\\ &= \sin x \times (-1)\\ &= -\sin x. \end{align} \]
What is the derivative of the function \(f(x) = 5\cos 3x + 2?\)
Differentiating \(f(x),\) we have
\[f'(x) = (5\cos 3x)' + (2)'. \]
Applying the chain rule gives
\[\begin{align} f'(x) &= 5\times(-\sin 3x)\times 3 + 0\\ &= -15\sin 3x. \ _\square \end{align}\]
Prove that the derivative of \(\tan x\) is \(\sec^2 x.\)
Using the quotient rule, we have
\[\begin{align} \tan x &= \frac{\sin x}{\cos x} \\\\ \Rightarrow (\tan x) ' &= \frac{\big(\cos x (\sin x)' - \sin x (\cos x)' \big)}{\cos^2 x} \\\\ &=\frac{\cos^2 x + \sin^2 x}{ \cos^2 x}\\\\ &=\frac{1}{\cos^2 x}\\\\ &= \sec^2 x. \ _\square \end{align}\]
The derivatives of the hyperbolic trigonometric functions are easier to compute.
By definition we know that
\[\sinh x = \frac{e^x - e^{-x}}{2} \qquad \text{and} \qquad \cosh x = \frac{e^x + e^{-x}}{2}.\]
Thus,
\[\begin{align} (\sinh x)' &= \frac{1}{2}\times \left( e^x - e^{-x}\right)' \\\\ &= \frac{1}{2}\big(e^x - (-e^x)\big)'\\\\ & = \frac{e^x+e^{-x}}{2} \\\\ &= \cosh x. \end{align}\]
If \(y = \sinh\big( \sin x + \cos(2\tan x)\big), \) what is its derivative \(y'\) with respect to \(x?\)
Using the chain rule, we get
\[\begin{align} y' &= \cosh\big(\sin x + \cos(2\tan x)\big) + \big(\sin x + \cos(2\tan x)\big)' \\\\ &= \cosh\big(\sin x + \cos(2\tan x)\big) + \cos x + \big(\cos(2\tan x)\big)'. \end{align}\]
Using the chain rule again, we have \(\cos(2\tan x)=-\sin(2\tan x)\times 2\sec^2 x. \) Therefore,
\[\begin{align} y'&= \cosh\big(\sin x + \cos(2\tan x)\big) + \cos x + \big(-\sin(2\tan x)\big)\times 2\sec^2 x \\\\ &= \cosh\big(\sin x + \cos(2\tan x)\big) + \cos x -2\sin(2\tan x)\sec^2 x. \ _\square \end{align} \]