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

$$\begin{aligned} f'(x) &= \lim_{h \rightarrow 0} \frac{\sin x \cos h + \sin h \cos x-\sin x}{h} \\\\ &= \lim_{h \rightarrow 0} \frac{\sin x (\cos h - 1) + \sin h \cos x}{h}. \end{aligned}$$

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 $x=\frac{\pi}{2}?$

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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

$$\begin{aligned} (\cos x)' &= \left(\sin\Big(\frac{\pi}{2} - x\Big)\right)' \times \left(\frac{\pi}{2} - x\right) ' \\ &= \cos\left(\frac{\pi}{2} - x\right) \times (- 1)\\ &= \sin x \times (-1)\\ &= -\sin x. \end{aligned}$$

What is the derivative of the function $f(x) = 5\cos 3x + 2?$

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Differentiating $f(x),$ we have

$$f'(x) = (5\cos 3x)' + (2)'.$$

Applying the chain rule gives

$$\begin{aligned} f'(x) &= 5\times(-\sin 3x)\times 3 + 0\\ &= -15\sin 3x. \ _\square \end{aligned}$$

Prove that the derivative of $\tan x$ is $\sec^2 x.$

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Using the quotient rule, we have

$$\begin{aligned} \tan x &= \frac{\sin x}{\cos x} \\\\ \Rightarrow (\tan x) ' &= \frac{\big(\cos x (\sin x)' - \sin x (\cos x)' \big)}{\cos^2 x} \\\\ &=\frac{\cos^2 x + \sin^2 x}{ \cos^2 x}\\\\ &=\frac{1}{\cos^2 x}\\\\ &= \sec^2 x. \ _\square \end{aligned}$$

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,

$$\begin{aligned} (\sinh x)' &= \frac{1}{2}\times \left( e^x - e^{-x}\right)' \\\\ &= \frac{1}{2}\big(e^x - (-e^x)\big)'\\\\ & = \frac{e^x+e^{-x}}{2} \\\\ &= \cosh x. \end{aligned}$$

If $y = \sinh\big( \sin x + \cos(2\tan x)\big),$ what is its derivative $y'$ with respect to $x?$

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Using the chain rule, we get

$$\begin{aligned} y' &= \cosh\big(\sin x + \cos(2\tan x)\big) + \big(\sin x + \cos(2\tan x)\big)' \\\\ &= \cosh\big(\sin x + \cos(2\tan x)\big) + \cos x + \big(\cos(2\tan x)\big)'. \end{aligned}$$

Using the chain rule again, we have $\cos(2\tan x)=-\sin(2\tan x)\times 2\sec^2 x.$ Therefore,

$$\begin{aligned} y'&= \cosh\big(\sin x + \cos(2\tan x)\big) + \cos x + \big(-\sin(2\tan x)\big)\times 2\sec^2 x \\\\ &= \cosh\big(\sin x + \cos(2\tan x)\big) + \cos x -2\sin(2\tan x)\sec^2 x. \ _\square \end{aligned}$$