Funções trigonométricas

Now that we know common techniques and limits in trigonometry, the limit definition of a derivative, and how to take the derivative of products and quotients, we can do all sorts of cool things. Specifically, we can find the derivatives of trigonometric functions, which will be very useful to us later.

Just to review, the derivative of a function $f$ at $x$, or $f'(x)$, is

$f'(x) = \lim_{h\to0} \frac{f(x+h) - f(x)}{h}.$

With that in mind, we set out to find the derivative of $\sin x$. Let's go back to our definition of a derivative. Now, our function $f$ is $\sin x$. So, our derivative, $\frac{d}{dx}\sin x$ is

$f'(x) = \lim_{h\to0} \frac{\sin(x+h) - \sin(x)}{h}.$

We can use our trusty angle sum formula! Now our limit becomes

$f'(x) = \lim_{h\to0} \frac{(\sin x\cos h + \sin h \cos x) - \sin(x)}{h}.$

We have lots of products in our limit. How can we factor those products into limits that should look familiar to us?

Our limit is equivalent to

$f'(x) = \sin x(\lim_{h\to0}\frac{\cos h - 1}{h}) + \cos x(\lim_{h\to0}\frac{\sin h}{h}).$

We remember (or derive quickly, or go back and learn how to) that

$\lim_{h\to0}\frac{\cos h - 1}{h} = 0$ and

$\lim_{h\to0}\frac{\sin h}{h} = 1.$

Substituting these values into our limit, we find that $f'(x) = \frac{d}{dx}\sin x = \cos x$! Nice. You'll probably want to remember that one.

Turns out that we can find the derivative of $\cos x$ in a very similar way.

Whew, we have accomplished a lot. We can now reap the benefits of our efforts. Remember that all of the basic trig functions can be written as combinations of sine and cosine? We can use the Quotient Rule to find all of their derivatives quickly! Let's walk through one example, and then you will be ready to conquer the rest.

We can compute that $\frac{d}{dx}\frac{\cos x}{\sin x} = \frac{\frac{d}{dx}(\cos x)(\sin x) - \frac{d}{dx}(\sin x)(\cos x)}{\sin^2 x}$ $= \frac{(-\sin x)(\sin x)-(\cos x)(\cos x)}{\sin^2 x} = \frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x}$ $= \frac{-1}{\sin^2 x} = -\csc^2 x.$

Awesome! Try $\tan x$, $\csc x$, and $\sec x$ for yourself before checking your answers below.

And that's all there is to it! (Well, until you run into those weirdos that like to multiply random trig functions together... but even then, you'll be able to handle it.)

Much like the proof of

$\frac{d}{dx}(\sin x) = \cos x$

we will prove that

$\frac{d}{dx}(\cos x) = -\sin x$

Use the definition of the derivative:

$\frac{d}{dx}(\cos x) = \lim_{h \to 0} \frac{\cos(x + h) - \cos(x)}{h}$

Use the angle addition formula for cosine:

$\lim_{h \to 0} \frac{\cos(x)\cos(h) - \sin(x)\sin(h) - \cos(x)}{h}$

Break into two limits:

$\lim_{h \to 0}\frac{\cos (x)\cos(h) - \cos(x)}{h} - \lim_{h \to 0}\frac{\sin(x)\sin(h)}{h}$

Rewrite:

$\cos(x)\lim_{h \to 0}\frac{\cos(h) - 1}{h} - \sin(x)\lim_{h \to 0}\frac{\sin(h)}{h}$

As cited in the previous note, we know that the first limit is $0$ and the second limit is $1$, so the expression simplifies to $-\sin(x)$, as desired.

Suppose we want to find the derivative of $\sin x$.

There are two approaches- I will explain the one that most people overlook. Remember that a derivative evaluated at $x=k$ gives the slope of the tangent line to the graph at $x=k$.

Lines with slope 0 are horizontal lines.

Let's look at the graph of $\sin x$. At $x= \pi, 3 \pi, 5\pi, ...$ it reaches the minimum/maximum values of $\sin x$.

At $x=\pi, 3\pi, 5\pi, ...$ the derivative of $\sin x$ must be 0, because the tangent lines will be horizontal.

Now let $f(x) = (\sin x)'$.

Then $f(x)$ must have zeroes at $x=\pi, x=3\pi, x=5\pi$, not to mention $x=-\pi, x=-3\pi, x=-5\pi, ...$

Therefore, $f(x) = a \cos x$ for some constant $a$. This is because $f(x)$ has roots in the exact same places as $\cos x$.

How do we find $a$? Let's take a look at $\sin x$ when x is close to 0.

Because $\sin x = - \sin (-x)$, as $x$ approaches 0, the slope of the tangent will approach $\frac{\sin x}{x}$ which approaches $1$.

Therefore, $f(0) = 1$, and we can find $a=1$.

We have shown that the derivative of $\sin x$ is $1 \cos x$, without using $\frac{f(x+h)-f(x)}{h}$

(Video) Complicated Derivative Examples

by PatrickJMT

Here's a video by patrickJMT going over some derivative problems that have a mix of different rules.

The first problem he looks at has a power rule, a chain rule, and a trig rule. $$y=\sqrt[3]{1+\tan x^4}$$ Rewrite this in order to remove the radical symbol. $$y=(1+\tan x^4)^{\frac13}$$ Now we can start to derive. $$\begin{align} y'&=\frac13(1+\tan x^4)^{-\frac23}\cdot\color{red}{[1+\tan x^4]'}\\ &=\frac13(1+\tan x^4)^{-\frac23}\cdot[\sec^2x^4\cdot\color{red}{(x^4)'}]\\ &=\frac13(1+\tan x^4)^{-\frac23}\cdot[\sec^2x^4\cdot(3x^4)]\\ &=\frac{3x^4\sec^2x^4}{3(1+\tan x^4)^{\frac23}}\\ &=\boxed{\frac{x^4\sec^2x^4}{(1+\tan x^4)^{\frac23}}} \end{align}$$

The next problem includes the chain rule and the product rule. $$\begin{align} y&=(t^4-1)^3(t^3+1)^4\\ y'&=\color{red}{[(t^4-1)^3]'}(t^3+1)^4+(t^4-1)^3\color{red}{[(t^3+1)^4]'}\\ &=[3(t^4-1)^2\color{red}{(t^4-1)'}](t^3+1)^4+(t^4-1)^3[4(t^3+1)^3\color{red}{(t^3+1)'}]\\ &=[3(t^4-1)^2(4t^3)](t^3+1)^4+(t^4-1)^3[4(t^3+1)^3(3t^2)]\\ &=12t^3(t^4-1)^2(t^3+1)^4+12t^2(t^4-1)^3(t^3+1)^3\\ &=12t^2(t^4-1)^2(t^3+1)^3[t(t^3+1)+(t^4-1)]\\ &=12t^2(t^4-1)^2(t^3+1)^3[t^4+t+t^4-1]\\ &=\boxed{12t^2(t^4-1)^2(t^3+1)^3(2t^4+t-1)} \end{align}$$

The final problem consists of the product rule, chain rule, and a trig rule. $$y=x\sin\frac1x$$ First let's rewrite fractions. $$\begin{align} y&=x\sin x^{-1}\\ y'&=\color{red}{(x)'}(\sin x^{-1})+(x)\color{red}{[\sin x^{-1}]'}\\ &=(1)(\sin x^{-1})+(x)[(\cos x^{-1})\color{red}{(x^{-1})'}]\\ &=(1)(\sin x^{-1})+(x)[(\cos x^{-1})(-1x^{-2})]\\ &=\boxed{\sin\frac1x-\frac{\cos\frac1x}{x}} \end{align}$$

(Video) Solving Every Derivative Ever

by PatrickJMT

Okay, maybe it's not EVERY derivative ever but here's a video by patrickJMT going over different derivatives for a whopping 35 minutes.

Let's have a season recap of some formulas before we go in. $$\begin{alignat}{2} & \frac{d}{dx}x^n=nx^{n-1} \quad && \frac{d}{dx}\cot x=-\csc^2 x\\ & \frac{d}{dx}k=0 \quad && \frac{d}{dx}\sec x=\sec x \tan x\\ & \frac{d}{dx}\sin x=\cos x \quad && \frac{d}{dx}e^x=e^x\\ & \frac{d}{dx}\cos x=-\sin x \quad && \frac{d}{dx}a^x=a^x\ln a\\ &\frac{d}{dx}\tan x=\sec^2x \quad && \frac{d}{dx}\ln x=\frac1x\\ &\frac{d}{dx}\csc x=-\csc x\cot x \quad &&\frac{d}{dx}\log_ax=\frac1{x\ln a} \end{alignat}$$

Also let's recap the product, quotient, and chain rule. $$\begin{align} [f(x)g(x)]'&=f'(x)g(x)+f(x)g'(x)\\ \bigg[\frac{f(x)}{g(x)}\bigg]'&=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}\\ [f(\ g(x)\ )]'&=f'(\ g(x)\ )\cdot g'(x) \end{align}$$

Problem 1

$$\begin{align} y&=x^2+3x+4\\ y'&=2x^{2-1}+3(1x^{1-1})+0\\ &=\boxed{2x+3} \end{align}$$

Problem 2

$$\begin{align} y&=5t^{-\frac23}\\ y'&=5(-\frac23t^{-\frac23-1})\\ &=-\frac{10}3t^{-\frac53}\\ &=\boxed{-\frac{10}{3t^{\frac53}}} \end{align}$$

Problem 3

$$\begin{align} y&=\frac{3x-1}{\sqrt x}\\ &=\frac{3x-1}{x^{\frac12}}\\ &=\frac{3x}{x^{\frac12}}-\frac{1}{x^{\frac12}}\\ &=3x^{\frac12}-x^{-\frac12}\\ y'&=3(\frac12x^{-\frac12})-(-\frac12x^{-\frac32})\\ &=\frac{3}{2x^{\frac12}}+\frac{1}{2x^{\frac32}}\\ &=\boxed{\frac{3x+1}{2x^{\frac32}}} \end{align}$$

Problem 4

$$\begin{align} y&=(x+4)(x-4)\\ &=x^2-4x+4x-16\\ &=x^2-16\\ y'&=2x^{2-1}-0\\ &=\boxed{2x} \end{align}$$

Problem 5

$$\begin{align} y&=\frac{\sqrt x-1}{\sqrt x+1}\\ &=\frac{x^{\frac12}-1}{x^{\frac12}+1}\\ y'&=\frac{\color{red}{(x^{\frac12}-1)'}(x^{\frac12}+1)-(x^{\frac12}-1)\color{red}{(x^{\frac12}+1)'}}{(x^{\frac12}+1)^2}\\ &=\frac{\frac12x^{-\frac12}(x^{\frac12}+1)-(x^{\frac12}-1)(\frac12x^{-\frac12})}{(x^{\frac12}+1)^2}\\ &=\frac{\frac12x^{-\frac12}[(x^{\frac12}+1)-(x^{\frac12}-1)]}{(x^{\frac12}+1)^2}\\ &=\frac{(x^{\frac12}+1)-(x^{\frac12}-1)}{2\sqrt x(x^{\frac12}+1)^2}\\ &=\frac{2}{2\sqrt x(x^{\frac12}+1)^2}\\ &=\boxed{\frac{1}{\sqrt x(x^{\frac12}+1)^2}} \end{align}$$

Problem 6

$$\begin{align} y&=\frac{2x-\sqrt{x}\cdot x}{x^2}\\ &=\frac{2x-x^{\frac32}}{x^2}\\ &=\frac{2x}{x^2}-\frac{x^{\frac32}}{x^2}\\ &=\frac{2}{x}-\frac{1}{x^{\frac12}}\\ &=2x^{-1}-x^{-\frac12}\\ y'&=2(-1x^{-2})-(-\frac12x^{\frac32})\\ &=\boxed{\frac{1}{2x^{\frac32}}-\frac{2}{x^2}} \end{align}$$

Problem 7

$$\begin{align} y&=(\sin x)(x^{\frac12})\\ y'&=\color{red}{(\sin x)'}(x^{\frac12})+(\sin x)\color{red}{(x^{\frac12})'}\\ &=(\cos x)(x^{\frac12})+(\sin x)(\frac12x^{-\frac12})\\ &=\boxed{\sqrt{x}\cos x+\frac{\sin x}{2\sqrt x}} \end{align}$$

Problem 8

$$\begin{align} y&=(\sin x)(\cos x)\\ y'&=\color{red}{(\sin x)'}(\cos x)+(\sin x)\color{red}{(\cos x)}\\ &=(\cos x)(\cos x)+(\sin x)(-\sin x)\\ &=\cos^2x-\sin^2x\\ &=\boxed{\cos(2x)} \end{align}$$

Problem 9

$$\begin{align} y&=\sin(\cos x)\\ y'&=\cos(\cos x)\cdot\color{red}{[\cos x]'}\\ &=\cos(\cos x)\cdot(-\sin x)\\ &=\boxed{-\sin x\cos(\cos x)} \end{align}$$

Problem 10

$$\begin{align} y&=\sec[(1+x^2)^{\frac12}]\\ y'&=\sec[(1+x^2)^{\frac12}]\cdot\tan[(1+x^2)^{\frac12}]\cdot\color{red}{[(1+x^2)^{\frac12}]'}\\ &=\sec[(1+x^2)^{\frac12}]\cdot\tan[(1+x^2)^{\frac12}]\cdot\frac12(1+x^2)^{-\frac12}\cdot\color{red}{(1+x^2)'}\\ &=\sec[(1+x^2)^{\frac12}]\cdot\tan[(1+x^2)^{\frac12}]\cdot\frac12(1+x^2)^{-\frac12}\cdot(2x)\\ &=\boxed{\frac{x\sec[(1+x^2)^{\frac12}]\tan[(1+x^2)^{\frac12}]}{\sqrt{1+x^2}}} \end{align}$$

Problem 11

$$\begin{align} y&=\tan^2[\sin(\theta^3)]\\ y'&=2\tan[\sin(\theta^3)]\cdot\color{red}{[\ \tan[\sin(\theta^3)]\ ]'}\\ &=2\tan[\sin(\theta^3)]\cdot\sec^2[\sin(\theta^3)]\cdot\color{red}{[\sin(\theta^3)]'}\\ &=2\tan[\sin(\theta^3)]\cdot\sec^2[\sin(\theta^3)]\cdot\cos(\theta^3)\cdot\color{red}{(\theta^3)'}\\ &=2\tan[\sin(\theta^3)]\cdot\sec^2[\sin(\theta^3)]\cdot\cos(\theta^3)\cdot3\theta^2\\ &=\boxed{6\theta^2\tan[\sin(\theta^3)]\sec^2[\sin(\theta^3)]\cos(\theta^3)} \end{align}$$

Problem 12

$$\begin{align} y&=x\sin(3x^{-1})\\ y'&=\color{red}{(x)'}[\sin(3x^{-1})]+(x)\color{red}{[\sin(3x^{-1})]'}\\ &=(1)\sin(3x^{-1})+(x)\cos(3x^{-1})\cdot\color{red}{(3x^{-1})'}\\ &=\sin(3x^{-1})+(x)\cos(3x^{-1})\cdot3(-1x^{-2})\\\\ &=\sin(3x^{-1})-\frac{3x\cos(3x^{-1})}{x^{2}}\\ &=\sin(\frac3x)-\frac{3\cos(\frac3x)}{x}\\ &=\boxed{\frac{x\sin(\frac3x)-3\cos(\frac3x)}{x}} \end{align}$$

Problem 13

$$\begin{align} y&=\frac{(x^3+4)^2}{(x^2+1)^4}\\ y'&=\frac{\color{red}{[(x^3+4)^2]'}(x^2+1)^4-(x^3+4)^2\color{red}{[(x^2+1)^4]'}}{[(x^2+1)^4]^2}\\ &=\frac{2(x^3+4)\color{red}{(x^3+4)'}(x^2+1)^4-4(x^3+4)^2(x^2+1)^3\color{red}{(x^2+1)'}}{(x^2+1)^8}\\ &=\frac{6x^2(x^3+4)(x^2+1)^4-8x(x^3+4)^2(x^2+1)^3}{(x^2+1)^8}\\ &=\frac{6x^2(x^3+4)(x^2+1)^4-8x(x^3+4)^2(x^2+1)^3}{(x^2+1)^8}\\ &=\frac{2x(x^3+4)(x^2+1)^3[3x(x^2+1)-4(x^3+4)]}{(x^2+1)^8}\\ &=\frac{2x(x^3+4)[3x(x^2+1)-4(x^3+4)]}{(x^2+1)^5}\\ &=\frac{2x(x^3+4)[(3x^3+3x)-(4x^3+16)]}{(x^2+1)^5}\\ &=\boxed{\frac{2x(x^3+4)(-x^3+3x-16)}{(x^2+1)^5}} \end{align}$$

Problem 14

$$\begin{align} y&=\ln(x)+x^3\\ y'&=\frac1x+3x^2\\ &=\boxed{\frac{3x^3+1}{x}} \end{align}$$

Problem 15

$$\begin{align} y&=x^2e^x\\ y'&=\color{red}{(x^2)'}e^x+x^2\color{red}{(e^x)'}\\ &=2xe^x+x^2e^x\\ &=\boxed{xe^x(x+2)} \end{align}$$

Problem 16

$$\begin{align} y&=e^{t\cdot\cos(2t)}\\ y'&=e^{t\cdot\cos(2t)}\cdot\color{red}{[t\cdot\cos(2t)]'}\\ &=e^{t\cdot\cos(2t)}\cdot[\color{red}{(t)'}\cos(2t)+t\color{red}{[\cos(2t)]'}]\\ &=e^{t\cdot\cos(2t)}\cdot[\cos(2t)-t\sin(2t)\cdot\color{red}{(2t)'}]\\ &=\boxed{e^{t\cdot\cos(2t)}\cdot[\cos(2t)-2t\sin(2t)]} \end{align}$$

Problem 17

$$\begin{align} y&=\cos(e^{\pi x})\\ y'&=-\sin(e^{\pi x})\cdot\color{red}{(e^{\pi x})'}\\ &=-\sin(e^{\pi x})\cdot e^{\pi x}\color{red}{(\pi x)'}\\ &=\boxed{-\pi e^{\pi x}\sin(e^{\pi x})} \end{align}$$

Problem 18

$$\begin{align} y&=(1+2e^{3x})^{\frac12}\\ y'&=\frac12(1+2e^{3x})^{-\frac12}\cdot\color{red}{(1+2e^{3x})'}\\ &=\frac12(1+2e^{3x})^{-\frac12}\cdot2e^{3x}\cdot\color{red}{(3x)'}\\ &=\boxed{\frac{3e^{3x}}{\sqrt{1+2e^{3x}}}} \end{align}$$

Problem 19

$$\begin{align} y&=5^{-\frac1x}\\ &=5^{-x^{-1}}\\ y'&=5^{-x^{-1}}\cdot\ln5\cdot\color{red}{(-x^{-1})'}\\ &=5^{-x^{-1}}\cdot\ln5\cdot(x^{-2})\\ &=\boxed{\frac{\ln5}{5^{\frac1x}x^2}} \end{align}$$

Problem 20

$$\begin{align} y&=\ln(x^4\sin^2x)\\ y'&=\frac{1}{x^4\sin^2x}\cdot\color{red}{[x^4\sin^2x]'}\\ &=\frac{1}{x^4\sin^2x}\cdot[\color{red}{(x^4)'}\sin^2x+x^4\color{red}{(\sin^2x)'}]\\ &=\frac{4x^3\sin^2x+2x^4\sin x\cdot\color{red}{(\sin x)'}}{x^4\sin^2x}\\ &=\frac{4x^3\sin^2x+2x^4\sin x\cos x}{x^4\sin^2x}\\ &=\frac{2x^3\sin x(2\sin x+x\cos x)}{x^4\sin^2x}\\ &=\frac{2(2+x\cos x)}{x\sin x}\\ &=\boxed{\frac4x+2\cot x} \end{align}$$

Problem 21

$$\begin{align} y&=\ln\sqrt{\frac{5x-1}{2x+3}}\\ &=\ln\bigg(\frac{5x-1}{2x+3}\bigg)^{\frac12}\\ &=\frac12\bigg[\ln\bigg(\frac{5x-1}{2x+3}\bigg)\bigg]\\ &=\frac12[\ln(5x-1)-\ln(2x+3)]\\ y'&=\frac12[\frac{1}{5x-1}\cdot\color{red}{(5x-1)'}-\frac{1}{2x+3}\cdot\color{red}{(2x+3)'}]\\ &=\frac12\bigg[\frac{5}{5x-1}-\frac{2}{2x+3}\bigg]\\ &=\frac12\bigg[\frac{5(2x+3)-2(5x-1)}{(5x-1)(2x+3)}\bigg]\\ &=\frac12\bigg[\frac{10x+15--10x+2}{10x^2+13x-3}\bigg]\\ &=\frac12\bigg[\frac{17}{10x^2+13x-3}\bigg]\\ &=\boxed{\frac{17}{20x^2+26x-6}} \end{align}$$