# Derivative of Trig Functions (Intuition Help?)

Looking for some intuition help here.

I have the following exercise and these are the steps I take: $$y = \sin\left(\frac{1}{x}\right)$$

$$u=\frac{1}{x}$$

$$y = \sin u,\;\;\frac{dy}{du} = \cos u= \cos\left(\frac{1}{x}\right)$$

$$u=x^{-1};\;\frac{du}{dx} =-x^{-2}=-\frac{1}{x^2}$$

$$\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}=cos\frac{1}{x}\times-\frac{1}{x^2}$$ This is incorrect but intuitively I want to multiply it this way. $$cos\frac{1}{x}\times-\frac{1}{x^2}=cos-\frac{1}{x^3}$$

But the correct answer is: $$-\frac{cos-\frac{1}{x}}{x^2}$$

Help me absorb the why so I can intuitively solve problems like these.

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Sorry, not sure I understand the question: you're struggling with the intuition behind calculating $\frac{d}{dx}\sin(\frac{1}{x})$? – Alex Wertheim May 12 '13 at 2:49
The intuition of multiplying cos1/x times -1/x^2 – ItsMitch May 12 '13 at 2:58

Here is where parentheses come in handy:

You found, correctly, $\dfrac{dy}{du}$ and $\dfrac{du}{dx}$.

But the scope of $\cos$ is restricted to its argument: $\left(\dfrac 1x\right)$ ONLY:

The FUNCTION $\dfrac{dy}{du} = \cos\left(\dfrac 1x\right)$ is multiplied by the function $\dfrac{du}{dx} = -\dfrac 1{x^2}$. That is not what you did. You multiplied argument of the $\cos$ function by the function $\dfrac{du}{dx} = -\dfrac{1}{x^2}$.

$$\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}=\left[cos\left(\frac{1}{x}\right)\right]\times\left(-\frac{1}{x^2}\right)$$ $$= -\frac 1{x^2} \cos\left(\frac 1x\right) = -\dfrac{\cos\left(\frac 1x\right)}{x^2}$$

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But why do I enclose it in brackets? – ItsMitch May 12 '13 at 2:56
Because $\cos$ is applied to the argument $\frac 1x$ ONLY. The FUNCTION $\dfrac{dy}{du} = \cos\left(\dfrac 1x\right)$ is multiplied by the function $\dfrac{du}{dx} = -\dfrac 1{x^2}$ – amWhy May 12 '13 at 3:03
Thank you very much! Where could I see more of these trig argument "laws" or rules? – ItsMitch May 12 '13 at 3:14
For example, If we have $f(x) = cos x,$ and we have $g(x) = x$, then $f(x)\cdot g(x) = \cos(x)\cdot (x) = x\cos(x)$ – amWhy May 12 '13 at 3:15
@amWhy: Even nice follw ups! +1 – Amzoti May 13 '13 at 0:23

First, you are taking short cuts when you are writing your equations that simply confuse things. Also, LaTeX has \sin and \cos. Start over.

\begin{aligned} \text{Let}\quad f(x) &= \sin\frac{1}{x}.\\ \text{Let}\quad u &= \frac{1}{x}.\\ f(x) &= \sin u\\ \frac{df}{dx} &= \frac{df}{du} \frac{du}{dx} \quad\text{by the Chain rule}\\ \frac{df}{du} &= \cos u\\ \frac{du}{dx} &= -\frac{1}{x^2} \\ \frac{df}{dx} &= \frac{df}{du} \frac{du}{dx} \\ &= \cos u \cdot [-\frac{1}{x^2}] \\ &= \cos \frac{1}{x} \cdot [-\frac{1}{x^2}]\\ &= -\frac{1}{x^2} \cos \frac{1}{x}. \end{aligned}

By the way, you are panicking and falling into the "Universal Distributive Law of Freshmen"; you think that all operations distribute over each other, and all operations commute. This is what has caused legions of students to replace $\sqrt{1+x^2}$ with $1 + x$. You wrote:

$$\cos\frac1x \cdot \frac{1}{x^2} = \cos\frac{1}{x^3}.$$

Even though $\cos x$ looks like a product, it isn't. Had you been using a programming language, you'd see that COS(X)*Y is not the same as COS(X*Y). Try any of these cases with actual numbers and a pocket calculator; you'll see what I mean.

When you panic like that, always take a deep breath and start over. And, never write a function next to its derivative and equate them.

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Thank you. Could you point me to a quick tutorial on LaTex and the correct way to write equations so I do not confuse more people? – ItsMitch May 12 '13 at 3:20
I should have looked at this too: meta.math.stackexchange.com/questions/107/… – Eric Jablow May 12 '13 at 3:24