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(2x^2tan(x))/sec(x) find f ' (x) and f ' (3)

Mathematics
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Answer is 2 x (x cos(x)+2 sin(x)), I just don't know how to do it
okay I got cha
it gets a little messy

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What do you do after you get the numerator?
Actually forget what I said. Just use the quotient rule
(4x)(tan(x))-(2x^2)(sec(x^2)) derivative of tan(x) = sec(x)^2
is that right for the numerator?
\[\frac{ \sec(x)(2x^{2}\sec^{2}x+4xtanx)- 2x^{2} \tan(x)\tan(x)\sec(x) }{ \sec^{2}x }\]
That's when it is in quotient rule form. As I said . it does get messy :/
Wow... sorry it took long to reply, the website isn't working well for me.
m trying to follow what you got and how you got it with the formula, i'm just learning the quotient rule
The quotient rule is low d(high) - high d(low) all over Low low or low^2
high is the numerator and low is the denominator when there is a d next to it that means the derivative
In this case the d(high) involved the product rule that's why it got so messy
okay im following now. Goodness this hard on the head... so first we found out what the numerator is from the product rule, and then we use the quotient rule
Just take it one step at a time :) Use the quotient rule for the whole thing. But you'll come across product rule inside it

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