anonymous
  • anonymous
find the equation of the tangent to the curve y=X^(x^2) - X^(Pi) where x=1 find the equation of the tangent to the curve y=X^(x^2) - X^(Pi) where x=1 @Mathematics
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Take log of y and then differentiate... (dy/dx) at x=1 will give the slope of the tangent...
anonymous
  • anonymous
when you say dy/dx at x=1 do you mean to plug in 1 for all x after i get the derivative of the equation?
anonymous
  • anonymous
exactly

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anonymous
  • anonymous
is the slop -10.6063
anonymous
  • anonymous
you can also find y at x=1... Thus you know the slope and one point in the line which will help you to find the equation...
anonymous
  • anonymous
what was your dy/dx
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=derivative+x%5E%28x%5E2%29-x%5E%28%CF%80%29
anonymous
  • anonymous
is it not (1-pi)?
anonymous
  • anonymous
yes yes thats the slope right? then how do i find the y to plug in for point slope form
anonymous
  • anonymous
you have the equation of the curve... put x=1 in it... that'll give you the point required...
anonymous
  • anonymous
that is y=X^(x^2) - X^(Pi) at x=1
anonymous
  • anonymous
= 0
anonymous
  • anonymous
yep
anonymous
  • anonymous
so you've got a point and the slope... go on...
anonymous
  • anonymous
can i leave it at y-0=1-pi(x-1)
anonymous
  • anonymous
y-0=(1-pi)(x-1) don't forget the brackets...
anonymous
  • anonymous
thats it?
anonymous
  • anonymous
you may write it in the form y=mx+c if you like...
anonymous
  • anonymous
y=-piX +X+pi-1
anonymous
  • anonymous
y=(1-pi)x-(1-pi)
anonymous
  • anonymous
thank you soo much!
anonymous
  • anonymous
you're welcome...
anonymous
  • anonymous
A piece of advice: you should've calculated the derivative by hand....
anonymous
  • anonymous
yea thats the problem i dont know how to do that one
anonymous
  • anonymous
do you want me to explain?
anonymous
  • anonymous
yes please
anonymous
  • anonymous
In y=X^(x^2) - X^(Pi) the problem is with the first term... I believe that you know how to differentiate x^(pi)...
anonymous
  • anonymous
yea
anonymous
  • anonymous
let y=p-q where p=X^(x^2) and q=X^(Pi)
anonymous
  • anonymous
taking log of p log(p)=x^2*log(x) then differentiate (1/p)dp/dx=x^2/x+2xlog(x) dp/dx=p[x+2xlog(x)] dp/dx=x^(x^2)*[x+2xlog(x)]
anonymous
  • anonymous
y=dp/dx-dq/dx
anonymous
  • anonymous
got it?
anonymous
  • anonymous
(1/p)dp/dx=x^2/x+2xlog(x)
anonymous
  • anonymous
so derivative of log(x) = 2xlog(x)
anonymous
  • anonymous
nope... derivative of log(x) is 1/x
anonymous
  • anonymous
p is the product of x^2 and log(x) so you've to use the product rule.
anonymous
  • anonymous
\[d(x^2*\log(x))/dx=x^2*d(\log(x))/dx + \log(x)*d(x^2)/dx\]
anonymous
  • anonymous
where is the 1/p from? (1/p)dp/dx=x^2/x+2xlog(x)
anonymous
  • anonymous
where did the x^2 go? dp/dx=p[x+2xlog(x)]
anonymous
  • anonymous
x^2/x=x

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