How to get (secx)(tanx) from (1/cosx)(sinx/cosx)?

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In summary: WarrenIn summary, the conversation discusses how to find the anti-derivative of a given function, specifically h(x) = sinx/cos^(2) x. The process involves converting the function into a simpler form, using substitution to find the anti-derivative, and recognizing that sin(x)/cos(x) is also known as tan(x) and 1/cos(x) is known as sec(x). The final answer is (secx)(tanx) + c.
  • #1
Quick question:

h(x) = sinx/cos^(2) x

= (1/cosx)(sinx/cosx)

Then you get (secx)(tanx)..

I do not get how they get secx x tanx?? Anyone?? Thanks
 
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  • #2
1/cos(x) is also called sec(x).

sin(x)/cos(x) is also called tan(x).

- Warren
 
  • #3
so the answer would be

(secx)(tanx) + c

Correct??
 
  • #4
All you've done so far is convert the function you gave me into a slightly simpler form.

sin(x) / cos^2(x) = sec(x) tan(x).

Since you didn't actually post the problem as it was given to you, I don't know if h(x) is a function of which you need to find the antiderivative, or whether you've already done that step. You probably need to actually perform the antiderivative now.

- Warren
 
  • #5
sorry how would i go about doing this.. Iam so confused.
 
  • #6
Find the function which has a derivative of sec(x) tan(x). You should have a list of such facts in your book.

- Warren
 
  • #7
alright.. I get it now.. Sorry.. I don't know why this problem was causing me problems.. Thanks for clearing it up.
 
  • #8
Do you know the method of substitution to find antiderivatives ? If so, just plug

[tex] \cos x = t [/tex]

and c what u get.

Daniel.
 
  • #9
It would have helped if you had told us from the beginning that you were trying to find an anti-derivative! All you said was that you couldn't see how they had gone from Quick question:

h(x) = sinx/cos^(2) x

to h(x)= (secx)(tanx).
 

What is the formula for (secx)(tanx)?

The formula for (secx)(tanx) is (1/cosx)(sinx/cosx).

How can I simplify (1/cosx)(sinx/cosx) to get (secx)(tanx)?

To simplify (1/cosx)(sinx/cosx), you can use the identity sinx/cosx = tanx. This will give you (1/cosx)(tanx), which is equivalent to (secx)(tanx).

Why is (secx)(tanx) equivalent to (1/cosx)(sinx/cosx)?

(secx)(tanx) is equivalent to (1/cosx)(sinx/cosx) because of the trigonometric identity secx = 1/cosx. This means that (secx)(tanx) can be rewritten as (1/cosx)(tanx), which can then be further simplified to (1/cosx)(sinx/cosx).

Can I use any other identities to get (secx)(tanx) from (1/cosx)(sinx/cosx)?

Yes, you can also use the identity secx = 1/cosx and tanx = sinx/cosx to get (secx)(tanx) from (1/cosx)(sinx/cosx). This will give you (1/cosx)(1/cosx)(sinx/cosx), which can be simplified to (1/cosx)(sinx/cosx).

Is there a visual representation of how (secx)(tanx) is derived from (1/cosx)(sinx/cosx)?

Yes, you can think of (secx)(tanx) as the product of two fractions: (1/cosx) and (sinx/cosx). When multiplied, the denominator of (1/cosx) and the numerator of (sinx/cosx) cancel out, leaving you with (1/cosx)(sinx/cosx) or (secx)(tanx).

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