anonymous
  • anonymous
find the derivative of y with respect to x, t or theta. (a) e^7 - 10x Answer: -10e^7-10x (b) 8xe^x - 8e^x answer: 8xe^x (c) y= (x^2 -2x+4)e^x answer: (x^2+2)e^x (d) y= sin e^theta^4 answer: (-4theta^3 e^-theta^4) cos e^-theta^4 please show the steps. thank you
Mathematics
schrodinger
  • schrodinger
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jim_thompson5910
  • jim_thompson5910
The first one is \[\Large e^{7-10x}\] right?
anonymous
  • anonymous
yes
jim_thompson5910
  • jim_thompson5910
The derivative of that isn't -10e^7-10x, so there has to be a typo somewhere

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anonymous
  • anonymous
im not sure i think we have to use the u substitution method to find the answer.
jim_thompson5910
  • jim_thompson5910
oh wait, is the answer is \[\Large -10e^{7-10x}\] and not \[\Large -10e^{7}-10x\]
anonymous
  • anonymous
yea its the first one
jim_thompson5910
  • jim_thompson5910
this is why parenthesis are useful if 7-10x is in the exponent, then say e^(7-10x)
jim_thompson5910
  • jim_thompson5910
ok that makes more sense now
anonymous
  • anonymous
sorry about that
jim_thompson5910
  • jim_thompson5910
let u = 7 - 10x, which means du/dx = -10 this means \[\Large e^{7-10x}\] turns into \[\Large e^{u}\] then derive to get \[\Large e^{u}*\frac{du}{dx}\] \[\Large e^{u}(-10)\] \[\Large -10e^{u}\] \[\Large -10e^{7-10x}\]
jim_thompson5910
  • jim_thompson5910
I'm using the chain rule, which is If h(x) = f(g(x)), then h ' (x) = f ' (g(x)) * g ' (x)
anonymous
  • anonymous
ok i just need help with the other 3 as well
jim_thompson5910
  • jim_thompson5910
whats the derivative of 8xe^x
anonymous
  • anonymous
im not sure
jim_thompson5910
  • jim_thompson5910
use the product rule
jim_thompson5910
  • jim_thompson5910
tell me what you get
anonymous
  • anonymous
is it 8xe^x
jim_thompson5910
  • jim_thompson5910
no, but that's part of it though
jim_thompson5910
  • jim_thompson5910
Product Rule: if h(x) = f(x)*g(x), then h ' (x) = f ' (x) * g(x) + f(x) * g ' (x)

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