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Math Help - sin cos tan derivatives

  1. #1
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    Exclamation sin cos tan derivatives

    Hey guys im stuck on 3 problems that are due in 2 days. I am hoping these problems are super easy for you guys to do so I can learn from the steps posted and ace my test 3 days from now. So here are the 3 problems.

    Use the chain rule (fog) = f(g(x)) to find the derivative dy/dx of each of the functions below.

    1. y= x*tan(square)x / 1+secx

    2. y=sin(sinx)


    3. y=cos(tanx)

    These are the problems that can help me past the test so please help me out.
    THANKS A LOT!!!
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  2. #2
    Senior Member Peritus's Avatar
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    tell us what is it that you don't understand.

    if you know how to apply the chain rule, and you are familiar with the the basic derivatives of the trigonometric functions (cos, sin tan...) there is no problem finding those derivatives.

    <br /> <br />
\frac{d}<br />
{{dx}}f\left( {g(x)} \right) = \left. {\frac{d}<br />
{{dt}}f(t)} \right|_{t = g(x)} \frac{d}<br />
{{dx}}g(x)<br /> <br />
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  3. #3
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    i know the rules but i don't know how to apply them to the problem to come up with the answer.
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  4. #4
    Senior Member Peritus's Avatar
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    Quote Originally Posted by uniquereason81 View Post
    i know the rules but i don't know how to apply them to the problem to come up with the answer.
    for exmple:

    2. f(x) = sinx, g(x) = sinx

    3. f(x) = cosx g(x) = tanx

    now can you solve 2 & 3 ?
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  5. #5
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    thanks Peritus but can u go futher with the problems. I understand how you got f(x) and g(x). THANKS
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  6. #6
    Senior Member Peritus's Avatar
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    Quote Originally Posted by uniquereason81 View Post
    thanks Peritus but can u go futher with the problems. I understand how you got f(x) and g(x). THANKS
    can you apply the chain rule:

    <br />
\frac{d}<br />
{{dx}}f\left( {g(x)} \right) = \left. {\frac{d}<br />
{{dt}}f(t)} \right|_{t = g(x)} \frac{d}<br />
{{dx}}g(x)<br />

    to f(x) and g(x) in 2& 3 ?
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  7. #7
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    no i can't apply it. I'm sorry I just learned this stuff last week.
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  8. #8
    Senior Member Peritus's Avatar
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    Quote Originally Posted by uniquereason81 View Post
    no i can't apply it. I'm sorry I just learned this stuff last week.
    ok, let us work problem 3 step by step:

    \begin{gathered}<br />
  f(x) = \cos (x) \hfill \\<br />
  g(x) = \tan (x) \hfill \\ <br />
\end{gathered}

    now let's look at the chain rule:

    <br />
\frac{d}<br />
{{dx}}f\left( {g(x)} \right) = \left. {\frac{d}<br />
{{dt}}f(t)} \right|_{t = g(x)} \frac{d}<br />
{{dx}}g(x)<br />

    first we have to differentiate f(x):


    f'(x) =  - \sin (x)

    now we have to substitute g(x) as the argument of this result:

    <br />
 - \sin (\tan x)<br />

    the next step is to calculate the derivative of g(x):

    <br />
g'(x) = \frac{1}<br />
{{\cos ^2 (x)}}

    the derivative of this composite function is the product of these two results:


    \frac{d}<br />
{{dx}}\cos (\tan x) = \frac{{ - \sin (\tan x)}}<br />
{{\cos ^2 (x)}}
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  9. #9
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    okay i think i got it so thanks but one question how do you know the inverses of sin cos and tan
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