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Math Help - Find the equation of the tangent line to the graph

  1. #16
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    Quote Originally Posted by pickslides View Post
    You can leave it as \displaystyle x=x_0 for a future value, then \displaystyle  y-f(x_0)=f'(x_0)(x-x_0)
    Hi zhangxupage2.

    Do as pickslides suggests. Substitute the following.

    f(x_0)=3x_0 \tan^2(x_0) + (x_0)^2 \cos(x_0)

    and as you found, f'(x_0)=3\tan^2(x_0)+6\tan(x_0)\sec^2(x_0) + 2x_0\cos(x_0)-(x_0)^2\sin(x_0)
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  2. #17
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    Thanks a lot everyone!
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  3. #18
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    What if the question asks me to pick a random value and plug it in, and the find the equation. What is an easy number to pick, considering that I can't use a calculator?
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  4. #19
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    Quote Originally Posted by zhangxupage2 View Post
    What if the question asks me to pick a random value and plug it in, and the find the equation. What is an easy number to pick, considering that I can't use a calculator?
    Values associated with the unit circle.
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