anonymous
  • anonymous
how do i find the derivative of: Y=x^3 tan^2 x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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myininaya
  • myininaya
you have a function times a function so you have to use the product rule
myininaya
  • myininaya
so we have y'=[x^3]'(tanx)^2+x^3[(tanx)^2]'
myininaya
  • myininaya
(x^3)'=3x^2 using power rule

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More answers

myininaya
  • myininaya
we have [(tanx)^2]'=2(tanx)*[secx]^2 using chain rule
anonymous
  • anonymous
thank you so much
myininaya
  • myininaya
np
anonymous
  • anonymous
\[3x^2(\tan^2x)+x^3(1/\cos^2x)\]
anonymous
  • anonymous
thank you
anonymous
  • anonymous
anytime :)
myininaya
  • myininaya
That's not right mmonish91
anonymous
  • anonymous
we know the equation is uv' + u'v, so take the following: - u = x^3 - u' = 3x^2 - v = tan^2x - v' = 2 tanx sec^2x so: y' = uv' + u'v = (x^3)(2tanxsec^2x) + (3x^2)(tan^2x)
anonymous
  • anonymous
^_^ simply as that.
anonymous
  • anonymous
simple*
anonymous
  • anonymous
@myinninaya please do explain
anonymous
  • anonymous
mmonish dear, you started correctly, but didn't get d/dx tan^2x right. the derivative of tanx = sec^2x , but in this case the derivative of: (tanx)^ 2 = 2.(tanx)(sec^2x) ; you multiply the power first then subtract it, then you derive tanx. (power rule) ^_^ a^ u = u(a)(u') <-- rule :)
anonymous
  • anonymous
i see...i left out the 2 for tan^2 mah bad
anonymous
  • anonymous
a^ u = u(a)^u-1 (u') , sorry lol
anonymous
  • anonymous
no problem, we all make mistakes ^_^
anonymous
  • anonymous
i don't really work with sec so my professor said we could leave it in sin or cos format
anonymous
  • anonymous
but you must know the derivative of tanx = sec^2x while the derivative of secx = secxtanx :)
anonymous
  • anonymous
which is same as 1/cos^2x
anonymous
  • anonymous
but I see what you mean :), u left it in cos, it's alright
anonymous
  • anonymous
yes, true :)
anonymous
  • anonymous
thank you guys!! ^.^
anonymous
  • anonymous
you are most welcome sindy ^_^ glad you understood it
anonymous
  • anonymous
yes i did understand :D ..

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