Calc

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KEYWEST17

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Jan 19, 2011
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Let F(x)= -3cosx+2tanx

F ' (x)=
F '(-pi/4)=

I simplified -3cosx to be 5sinx
Not sure how to simplify 2 tanx
 
Are you sure you understand the assignment?

\(\displaystyle F(x) = -3\cos(x)+2\tan(x)\)

\(\displaystyle F'(x) = 3\sin(x)+2\sec^{2}(x)\)

Now what?
 
Can you show me step by step how you arrived at that answer and if you have time walk me through how to get F '(-pi/4)
 
That was the first step and the ONLY step that had anything to do with the calculus.

You simply must know the derivatives of sine, cosine, and tangent. If you do not know them, you will fail.

The last part is a simple substitution. You should be able to do that. With a standard value like -pi/4, these values should be on the tip of your tongue or you may need to reconsider whether you belong in calculus class.

Let's see what you get.
 
1) Please give up on the degrees thing. This will not get you through the course. You need that, but you also need a comprehension of the function defininition, independent of the unit circle.

2) Why would the sine or cosine function produce a result indegrees?
 
What am I missing then to finish the problem? Somewhere I am missing a step and don't know my mistake. I would appreciate if you could help me out.
 
Substitution of the given value into the function.
 
I think the answer of F'(-pi/4) would be:

F'(x)=3(-sqrt2/2)+2(sqrt2)^2, just simplify that and you should have the answer. You are just plugging -pi/4 in for x. sin(-pi/4) is -sqrt2/2 and sec(-pi/4) is sqrt2. I'm in pre calc and we haven't studied the derivatives of trig functions yet so I'm assuming what tkhunny put down for F'(x) is correct.
 
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