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Ok so I think my book is really not explaining this properly. Here's what I have so far:

Derivative of $$\sin x \text{ is } \cos x.$$ Because the derivative is the negative sine function, which would then be a cosine function. Ok, makes perfect sense.

Derivative of $$\tan x \text{ is }\sec^2x.$$

Okay, great. But what's stumping me is I have this equation $$\sin(x) \tan(x).$$ On my homework, (it's a software so I know if I got it right) I got it right, but I can't for the life of me understand why.

The answer comes out to $$\sin(x)\sec^2(x) + \sin(x)$$ and in my book they don't explain any of this at all. They just want me to accept it but I need to understand why.

Wouldn't it be $$\cos(x)\sec^2(x) + \cos(x)\text{ for }f(x + h) + f(x) ?$$

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    $\begingroup$ Are you familiar with the product rule? $\endgroup$ – Esteemator Mar 7 '15 at 0:56
  • $\begingroup$ Yes I'm very familiar with it. I don't see how the product rule gives us that. $\endgroup$ – Omeed Mar 7 '15 at 1:04
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    $\begingroup$ You seem confused about several things. You refer to "the negative sine function, which would then be a cosine function". The cosine function is not "the negative sine function". Then you say "I have this equation". What follows is not an equation. You could say "I have this function". You refer to $f(x+h)+f(x)$ for no particular reason. The expression $\dfrac{f(x+h)-f(x)} h$ is the slope of a secant line and its limits as $h$ approaches $0$ is $f'(x)$, but why you mention $f(x+h)+f(x)$ is a mystery. ${}\qquad{}$ $\endgroup$ – Michael Hardy Mar 7 '15 at 3:39
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You get the correct answer when you use the product rule

Differentiate one function, hold the rest constant. Repeat for each term. $$(uv)' = \underbrace{uv'}_{u \text{ held constant}} + \underbrace{vu'}_{v\text{ held constant}}$$

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  • $\begingroup$ $$sin(x)sec^2x + tan(x)cos(x)$$, so how does the answer come out to $$sin(x)sec^2(x) + sin(x)$$ ?? $\endgroup$ – Omeed Mar 7 '15 at 1:05
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    $\begingroup$ @Omeed $\tan x = \frac{\sin x}{\cos x}$ $\endgroup$ – Jack Mar 7 '15 at 1:07
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    $\begingroup$ @Omeed: $\tan(x)\cos(x) = \frac{\sin(x)}{\cos(x)} \cos(x) = \sin(x)$ $\endgroup$ – Henry Mar 7 '15 at 1:10

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